✅Beats 100% | Video | Explained Step by Step | 2 Approaches
https://youtu.be/eHZLQIH1ruk?si=Z9939JPiMUU2OJFQ\n\n\n# #1 Two Pointers + Greedy\n\npython3 []\nclass Solution:\n    def findLengthOfShortestSubarray(self, arr:
https://youtu.be/eHZLQIH1ruk?si=Z9939JPiMUU2OJFQ\n\n\n# #1 Two Pointers + Greedy\n\n```python3 []\nclass Solution:\n    def findLengthOfShortestSubarray(self, arr: List[int]) -> int:\n        n = len(arr)\n        \n        # 1. Find the rightmost point where array becomes non-ascending\n        right = n - 1\n        while right > 0 and arr[right - 1] <= arr[right]:\n            right -= 1\n            \n        # 2. If array is already sorted in ascending order\n        if right == 0:\n            return 0\n            \n        # 3. Initial minimum length is from start to right pointer\n        min_length = right\n        \n        # 4. Check each element from left side\n        for left in range(n):\n            # 4.1 Break if left sequence becomes non-ascending\n            if left > 0 and arr[left - 1] > arr[left]:\n                break\n                \n            # 4.2 Find the first element from right that\'s >= arr[left]\n            while right < n and arr[left] > arr[right]:\n                right += 1\n                \n            # 4.3 Update minimum length of subarray to be removed\n            current_length = right - left - 1\n            min_length = min(min_length, current_length)\n            \n        return min_length\n```\n```cpp []\nclass Solution {\npublic:\n    int findLengthOfShortestSubarray(vector<int>& arr) {\n        int n = arr.size();\n        \n        // 1.Find the rightmost point where array becomes non-ascending\n        int right = n - 1;\n        while (right > 0 && arr[right - 1] <= arr[right]) {\n            right--;\n        }\n        \n        // 2.If array is already sorted in ascending order\n        if (right == 0) {\n            return 0;\n        }\n        \n        // 3.Initial minimum length is from start to right pointer\n        int min_length = right;\n        \n        // Check each element from left side\n        for (int left = 0; left < n; left++) {\n            // Break if left sequence becomes non-ascending\n            if (left > 0 && arr[left - 1] > arr[left]) {\n                break;\n            }\n            \n            // Find the first element from right that\'s >= arr[left]\n            while (right < n && arr[left] > arr[right]) {\n                right++;\n            }\n            \n            // Update minimum length of subarray to be removed\n            int current_length = right - left - 1;\n            min_length = min(min_length, current_length);\n        }\n        \n        return min_length;\n    }\n};\n```\n```java []\nclass Solution {\n    public int findLengthOfShortestSubarray(int[] arr) {\n        int n = arr.length;\n        \n        // 1.Find the rightmost point where array becomes non-ascending\n        int right = n - 1;\n        while (right > 0 && arr[right - 1] <= arr[right]) {\n            right--;\n        }\n        \n        // 2.If array is already sorted in ascending order\n        if (right == 0) {\n            return 0;\n        }\n        \n        // 3.Initial minimum length is from start to right pointer\n        int minLength = right;\n        \n        // Check each element from left side\n        for (int left = 0; left < n; left++) {\n            // Break if left sequence becomes non-ascending\n            if (left > 0 && arr[left - 1] > arr[left]) {\n                break;\n            }\n            \n            // Find the first element from right that\'s >= arr[left]\n            while (right < n && arr[left] > arr[right]) {\n                right++;\n            }\n            \n            // Update minimum length of subarray to be removed\n            int currentLength = right - left - 1;\n            minLength = Math.min(minLength, currentLength);\n        }\n        \n        return minLength;\n    }\n}\n```\n```csharp []\npublic class Solution {\n    public int FindLengthOfShortestSubarray(int[] arr) {\n        int n = arr.Length;\n        \n        // 1. Find the rightmost point where array becomes non-ascending\n        int right = n - 1;\n        while (right > 0 && arr[right - 1] <= arr[right]) {\n            right--;\n        }\n        \n        // 2. If array is already sorted in ascending order\n        if (right == 0) {\n            return 0;\n        }\n        \n        // 3. Initial minimum length is from start to right pointer\n        int minLength = right;\n        \n        // Check each element from left side\n        for (int left = 0; left < n; left++) {\n            // Break if left sequence becomes non-ascending\n            if (left > 0 && arr[left - 1] > arr[left]) {\n                break;\n            }\n            \n            // Find the first element from right that\'s >= arr[left]\n            while (right < n && arr[left] > arr[right]) {\n                right++;\n            }\n            \n            // Update minimum length of subarray to be removed\n            int currentLength = right - left - 1;\n            minLength = Math.Min(minLength, currentLength);\n        }\n        \n        return minLength;\n    }\n}\n```\n```golang []\nfunc findLengthOfShortestSubarray(arr []int) int {\n    n := len(arr)\n    \n    // 1. Find the rightmost point where array becomes non-ascending\n    right := n - 1\n    for right > 0 && arr[right-1] <= arr[right] {\n        right--\n    }\n    \n    // 2. If array is already sorted in ascending order\n    if right == 0 {\n        return 0\n    }\n    \n    // 3. Initial minimum length is from start to right pointer\n    minLength := right\n    \n    // Check each element from left side\n    for left := 0; left < n; left++ {\n        // Break if left sequence becomes non-ascending\n        if left > 0 && arr[left-1] > arr[left] {\n            break\n        }\n        \n        // Find the first element from right that\'s >= arr[left]\n        for right < n && arr[left] > arr[right] {\n            right++\n        }\n        \n        // Update minimum length of subarray to be removed\n        currentLength := right - left - 1\n        minLength = min(minLength, currentLength)\n    }\n    \n    return minLength\n}\n```\n```swift []\nclass Solution {\n    func findLengthOfShortestSubarray(_ arr: [Int]) -> Int {\n        let n = arr.count\n        \n        // 1. Find the rightmost point where array becomes non-ascending\n        var right = n - 1\n        while right > 0 && arr[right - 1] <= arr[right] {\n            right -= 1\n        }\n        \n        // 2. If array is already sorted in ascending order\n        if right == 0 {\n            return 0\n        }\n        \n        // 3. Initial minimum length is from start to right pointer\n        var minLength = right\n        \n        // Check each element from left side\n        for left in 0..<n {\n            // Break if left sequence becomes non-ascending\n            if left > 0 && arr[left - 1] > arr[left] {\n                break\n            }\n            \n            // Find the first element from right that\'s >= arr[left]\n            while right < n && arr[left] > arr[right] {\n                right += 1\n            }\n            \n            // Update minimum length of subarray to be removed\n            let currentLength = right - left - 1\n            minLength = min(minLength, currentLength)\n        }\n        \n        return minLength\n    }\n}\n```\n```javascript [JS]\n// JavaScript\n\nvar findLengthOfShortestSubarray = function(arr) {\n    const n = arr.length;\n    \n    // 1. Find the rightmost point where array becomes non-ascending\n    let right = n - 1;\n    while (right > 0 && arr[right - 1] <= arr[right]) {\n        right--;\n    }\n    \n    // 2. If array is already sorted in ascending order\n    if (right === 0) {\n        return 0;\n    }\n    \n    // 3. Initial minimum length is from start to right pointer\n    let minLength = right;\n    \n    // Check each element from left side\n    for (let left = 0; left < n; left++) {\n        // Break if left sequence becomes non-ascending\n        if (left > 0 && arr[left - 1] > arr[left]) {\n            break;\n        }\n        \n        // Find the first element from right that\'s >= arr[left]\n        while (right < n && arr[left] > arr[right]) {\n            right++;\n        }\n        \n        // Update minimum length of subarray to be removed\n        const currentLength = right - left - 1;\n        minLength = Math.min(minLength, currentLength);\n    }\n    \n    return minLength;\n};\n```\n```typescript [TS]\n// TypeScript\n\nfunction findLengthOfShortestSubarray(arr: number[]): number {\n    const n: number = arr.length;\n    \n    // 1. Find the rightmost point where array becomes non-ascending\n    let right: number = n - 1;\n    while (right > 0 && arr[right - 1] <= arr[right]) {\n        right--;\n    }\n    \n    // 2. If array is already sorted in ascending order\n    if (right === 0) {\n        return 0;\n    }\n    \n    // 3. Initial minimum length is from start to right pointer\n    let minLength: number = right;\n    \n    // Check each element from left side\n    for (let left = 0; left < n; left++) {\n        // Break if left sequence becomes non-ascending\n        if (left > 0 && arr[left - 1] > arr[left]) {\n            break;\n        }\n        \n        // Find the first element from right that\'s >= arr[left]\n        while (right < n && arr[left] > arr[right]) {\n            right++;\n        }\n        \n        // Update minimum length of subarray to be removed\n        const currentLength: number = right - left - 1;\n        minLength = Math.min(minLength, currentLength);\n    }\n    \n    return minLength;\n}\n```\n```rust []\nimpl Solution {\n    pub fn find_length_of_shortest_subarray(arr: Vec<i32>) -> i32 {\n        let n = arr.len();\n        \n        // 1. Find the rightmost point where array becomes non-ascending\n        let mut right = n - 1;\n        while right > 0 && arr[right - 1] <= arr[right] {\n            right -= 1;\n        }\n        \n        // 2. If array is already sorted in ascending order\n        if right == 0 {\n            return 0;\n        }\n        \n        // 3. Initial minimum length is from start to right pointer\n        let mut min_length = right;\n        \n        // Check each element from left side\n        for left in 0..n {\n            // Break if left sequence becomes non-ascending\n            if left > 0 && arr[left - 1] > arr[left] {\n                break;\n            }\n            \n            // Find the first element from right that\'s >= arr[left]\n            while right < n && arr[left] > arr[right] {\n                right += 1;\n            }\n            \n            // Update minimum length of subarray to be removed\n            let current_length = right - left - 1;\n            min_length = min_length.min(current_length);\n        }\n        \n        min_length as i32\n    }\n}\n```\n```ruby []\ndef find_length_of_shortest_subarray(arr)\n    n = arr.length\n    \n    # 1. Find the rightmost point where array becomes non-ascending\n    right = n - 1\n    while right > 0 && arr[right - 1] <= arr[right]\n        right -= 1\n    end\n    \n    # 2. If array is already sorted in ascending order\n    return 0 if right == 0\n    \n    # 3. Initial minimum length is from start to right pointer\n    min_length = right\n    \n    # Check each element from left side\n    (0...n).each do |left|\n        # Break if left sequence becomes non-ascending\n        break if left > 0 && arr[left - 1] > arr[left]\n        \n        # Find the first element from right that\'s >= arr[left]\n        while right < n && arr[left] > arr[right]\n            right += 1\n        end\n        \n        # Update minimum length of subarray to be removed\n        current_length = right - left - 1\n        min_length = [min_length, current_length].min\n    end\n    \n    min_length\nend\n```\n\n- Complexy: Time O(n) & Space O(1)\n\n# #2 Two Pointers + Sliding Window (from video but fixed, all beats 100%)\n\n\n\n\n```python3 []\nclass Solution:\n    def findLengthOfShortestSubarray(self, arr: List[int]) -> int:\n        n = len(arr)\n        if n == 1:\n            return 0\n            \n        # Find the longest non-decreasing prefix\n        left = 0\n        while left + 1 < n and arr[left] <= arr[left + 1]:\n            left += 1\n            \n        if left == n - 1:  # Array is already non-decreasing\n            return 0\n            \n        # Find the longest non-decreasing suffix\n        right = n - 1\n        while right > 0 and arr[right - 1] <= arr[right]:\n            right -= 1\n            \n        # Initial result: remove everything between left and right\n        result = min(n - left - 1, right)\n        \n        # Try to merge prefix and suffix\n        i, j = 0, right\n        while i <= left and j < n:\n            if arr[i] <= arr[j]:\n                result = min(result, j - i - 1)\n                i += 1\n            else:\n                j += 1\n                \n        return result\n\n```\n```cpp []\nclass Solution {\npublic:\n    int findLengthOfShortestSubarray(vector<int>& arr) {\n        int n = arr.size();\n        \n        if (n == 1) {\n            return 0;\n        }\n        \n        // Find the longest non-decreasing prefix\n        int left = 0;\n        while (left + 1 < n && arr[left] <= arr[left + 1]) {\n            left++;\n        }\n        \n        if (left == n - 1) {  // Array is already non-decreasing\n            return 0;\n        }\n        \n        // Find the longest non-decreasing suffix\n        int right = n - 1;\n        while (right > 0 && arr[right - 1] <= arr[right]) {\n            right--;\n        }\n        \n        // Initial result: remove everything between left and right\n        int result = min(n - left - 1, right);\n        \n        // Try to merge prefix and suffix\n        int i = 0, j = right;\n        while (i <= left && j < n) {\n            if (arr[i] <= arr[j]) {\n                result = min(result, j - i - 1);\n                i++;\n            } else {\n                j++;\n            }\n        }\n        \n        return result;\n    }\n};\n```\n```java []\nclass Solution {\n    public int findLengthOfShortestSubarray(int[] arr) {\n        int n = arr.length;\n        \n        if (n == 1) {\n            return 0;\n        }\n        \n        // Find the longest non-decreasing prefix\n        int left = 0;\n        while (left + 1 < n && arr[left] <= arr[left + 1]) {\n            left++;\n        }\n        \n        if (left == n - 1) {  // Array is already non-decreasing\n            return 0;\n        }\n        \n        // Find the longest non-decreasing suffix\n        int right = n - 1;\n        while (right > 0 && arr[right - 1] <= arr[right]) {\n            right--;\n        }\n        \n        // Initial result: remove everything between left and right\n        int result = Math.min(n - left - 1, right);\n        \n        // Try to merge prefix and suffix\n        int i = 0, j = right;\n        while (i <= left && j < n) {\n            if (arr[i] <= arr[j]) {\n                result = Math.min(result, j - i - 1);\n                i++;\n            } else {\n                j++;\n            }\n        }\n        \n        return result;\n    }\n}\n```\n```csharp []\npublic class Solution {\n    public int FindLengthOfShortestSubarray(int[] arr) {\n        int n = arr.Length;\n        if (n == 1) return 0;\n        \n        // Find the longest non-decreasing prefix\n        int left = 0;\n        while (left + 1 < n && arr[left] <= arr[left + 1]) {\n            left++;\n        }\n        \n        if (left == n - 1) return 0;\n        \n        // Find the longest non-decreasing suffix\n        int right = n - 1;\n        while (right > 0 && arr[right - 1] <= arr[right]) {\n            right--;\n        }\n        \n        // Initial result: remove everything between left and right\n        int result = Math.Min(n - left - 1, right);\n        \n        // Try to merge prefix and suffix\n        int i = 0, j = right;\n        while (i <= left && j < n) {\n            if (arr[i] <= arr[j]) {\n                result = Math.Min(result, j - i - 1);\n                i++;\n            } else {\n                j++;\n            }\n        }\n        \n        return result;\n    }\n}\n```\n```golang []\nfunc findLengthOfShortestSubarray(arr []int) int {\n    n := len(arr)\n    if n == 1 {\n        return 0\n    }\n    \n    // Find the longest non-decreasing prefix\n    left := 0\n    for left+1 < n && arr[left] <= arr[left+1] {\n        left++\n    }\n    \n    if left == n-1 {\n        return 0\n    }\n    \n    // Find the longest non-decreasing suffix\n    right := n - 1\n    for right > 0 && arr[right-1] <= arr[right] {\n        right--\n    }\n    \n    // Initial result: remove everything between left and right\n    result := min(n-left-1, right)\n    \n    // Try to merge prefix and suffix\n    i, j := 0, right\n    for i <= left && j < n {\n        if arr[i] <= arr[j] {\n            result = min(result, j-i-1)\n            i++\n        } else {\n            j++\n        }\n    }\n    \n    return result\n}\n\nfunc min(a, b int) int {\n    if a < b {\n        return a\n    }\n    return b\n}\n```\n```swift []\nclass Solution {\n    func findLengthOfShortestSubarray(_ arr: [Int]) -> Int {\n        let n = arr.count\n        if n == 1 { return 0 }\n        \n        // Find the longest non-decreasing prefix\n        var left = 0\n        while left + 1 < n && arr[left] <= arr[left + 1] {\n            left += 1\n        }\n        \n        if left == n - 1 { return 0 }\n        \n        // Find the longest non-decreasing suffix\n        var right = n - 1\n        while right > 0 && arr[right - 1] <= arr[right] {\n            right -= 1\n        }\n        \n        // Initial result: remove everything between left and right\n        var result = min(n - left - 1, right)\n        \n        // Try to merge prefix and suffix\n        var i = 0, j = right\n        while i <= left && j < n {\n            if arr[i] <= arr[j] {\n                result = min(result, j - i - 1)\n                i += 1\n            } else {\n                j += 1\n            }\n        }\n        \n        return result\n    }\n}\n```\n```javascript [JS]\n// JavaScript\n\nvar findLengthOfShortestSubarray = function(arr) {\n    const n = arr.length;\n    if (n === 1) return 0;\n    \n    // Find the longest non-decreasing prefix\n    let left = 0;\n    while (left + 1 < n && arr[left] <= arr[left + 1]) {\n        left++;\n    }\n    \n    if (left === n - 1) return 0;\n    \n    // Find the longest non-decreasing suffix\n    let right = n - 1;\n    while (right > 0 && arr[right - 1] <= arr[right]) {\n        right--;\n    }\n    \n    // Initial result: remove everything between left and right\n    let result = Math.min(n - left - 1, right);\n    \n    // Try to merge prefix and suffix\n    let i = 0, j = right;\n    while (i <= left && j < n) {\n        if (arr[i] <= arr[j]) {\n            result = Math.min(result, j - i - 1);\n            i++;\n        } else {\n            j++;\n        }\n    }\n    \n    return result;\n};\n```\n```typescript [TS]\n// TypeScript\n\nfunction findLengthOfShortestSubarray(arr: number[]): number {\n    const n: number = arr.length;\n    if (n === 1) return 0;\n    \n    // Find the longest non-decreasing prefix\n    let left: number = 0;\n    while (left + 1 < n && arr[left] <= arr[left + 1]) {\n        left++;\n    }\n    \n    if (left === n - 1) return 0;\n    \n    // Find the longest non-decreasing suffix\n    let right: number = n - 1;\n    while (right > 0 && arr[right - 1] <= arr[right]) {\n        right--;\n    }\n    \n    // Initial result: remove everything between left and right\n    let result: number = Math.min(n - left - 1, right);\n    \n    // Try to merge prefix and suffix\n    let i: number = 0, j: number = right;\n    while (i <= left && j < n) {\n        if (arr[i] <= arr[j]) {\n            result = Math.min(result, j - i - 1);\n            i++;\n        } else {\n            j++;\n        }\n    }\n    \n    return result;\n}\n```\n```rust []\nimpl Solution {\n    pub fn find_length_of_shortest_subarray(arr: Vec<i32>) -> i32 {\n        let n = arr.len();\n        if n == 1 {\n            return 0;\n        }\n        \n        // Find the longest non-decreasing prefix\n        let mut left = 0;\n        while left + 1 < n && arr[left] <= arr[left + 1] {\n            left += 1;\n        }\n        \n        if left == n - 1 {\n            return 0;\n        }\n        \n        // Find the longest non-decreasing suffix\n        let mut right = n - 1;\n        while right > 0 && arr[right - 1] <= arr[right] {\n            right -= 1;\n        }\n        \n        // Initial result: remove everything between left and right\n        let mut result = std::cmp::min(n - left - 1, right);\n        \n        // Try to merge prefix and suffix\n        let mut i = 0;\n        let mut j = right;\n        while i <= left && j < n {\n            if arr[i] <= arr[j] {\n                result = std::cmp::min(result, j - i - 1);\n                i += 1;\n            } else {\n                j += 1;\n            }\n        }\n        \n        result as i32\n    }\n}\n```\n```ruby []\ndef find_length_of_shortest_subarray(arr)\n    n = arr.length\n    return 0 if n == 1\n    \n    # Find the longest non-decreasing prefix\n    left = 0\n    left += 1 while left + 1 < n && arr[left] <= arr[left + 1]\n    \n    return 0 if left == n - 1\n    \n    # Find the longest non-decreasing suffix\n    right = n - 1\n    right -= 1 while right > 0 && arr[right - 1] <= arr[right]\n    \n    # Initial result: remove everything between left and right\n    result = [n - left - 1, right].min\n    \n    # Try to merge prefix and suffix\n    i, j = 0, right\n    while i <= left && j < n\n        if arr[i] <= arr[j]\n            result = [result, j - i - 1].min\n            i += 1\n        else\n            j += 1\n        end\n    end\n    \n    result\nend\n```\n\n\n\n\n- Complexy: Time O(n) & Space O(1)\n\n# Explanation\n---\n\n## Aproach #2 Two Pointers + Sliding Window:\n\n## video above /\\\n\n\n## Aproach #1 Two Pointers + Greedy\n\n\n1. First, we find the rightmost point where the array stops being sorted (non-decreasing). We start from the end and move left until we find a point where the order breaks.\n\n```\nclass Solution:\n    def findLengthOfShortestSubarray(self, arr: List[int]) -> int:\n        n = len(arr)\n        \n        # Step 1: Find rightmost point where array becomes non-ascending\n        right = n - 1\n        while right > 0 and arr[right - 1] <= arr[right]:\n            right -= 1\n```\n\n2. If right pointer reaches 0, it means the entire array is already sorted, so we return 0 (no removal needed).\n```\n        # Step 2: Check if array is already sorted\n        if right == 0:\n            return 0\n```\n3. Initially, we set the minimum length as \'right\' - assuming we remove everything from start to \'right\'.\n\n```\n        # Step 3: Initialize minimum length\n        min_length = right\n```\n4. Main loop:\n    - Iterate through array from left\n    - For each valid left point (maintaining ascending order)\n    - Find the rightmost point where we can connect left and right parts\n    - Calculate current subarray length to remove\n    - Update minimum length if curent is smaller\n\n```\n        # Step 4: Main loop to find minimum length\n        for left in range(n):\n            # Break if left sequence becomes non-ascending\n            if left > 0 and arr[left - 1] > arr[left]:\n                break\n                \n            # Find first element from right that\'s >= arr[left]\n            while right < n and arr[left] > arr[right]:\n                right += 1\n                \n            # Update minimum length\n            current_length = right - left - 1\n            min_length = min(min_length, current_length)\n```\n\n### Core Logic:\n\n- Finding a sorted suffix from the right\n- Finding a sorted prefix from the left\n- For each valid prefix, finding the shortest middle part that needs to be removed to connect prefix and suffix while maintaining sorted order\n\nDone.\n\n---\n![array2.png](https://assets.leetcode.com/users/images/7a0a6937-9d07-4cf6-ad93-d27ffcaf69a0_1730577425.4474297.png)\n\n# Most common Array Interview Problems\nList based on my research (interviews, reddit, github) Probably not 100% accurate but very close to my recruitments. Leetcode Premium is probably more accurate [I don\'t have]\n\n\n**Easy:** [[1. Two Sum]](https://leetcode.com/problems/two-sum/solutions/5999466/beats-100-explained-step-by-step-list-most-common-array-inverview) **(common warm-up)** [[121. Best Time to Buy and Sell Stock]](https://leetcode.com/discuss/topic/6027720/solution/) [[88. Merge Sorted Array]](https://leetcode.com/problems/merge-sorted-array/solutions/6031394/beats-100-explained-step-by-step-list-most-common-string-interview-problems) [[27. Remove Element]](https://leetcode.com/problems/remove-element/solutions/6031528/beats-100-explained-step-by-step-list-most-common-array-interview-problems) [[929. Unique Email Addresses]](https://leetcode.com/problems/unique-email-addresses/solutions/6031687/explained-step-by-step-video-list-most-common-array-interview-problems) \n\n\n**Medium/Hard:** [[15. 3Sum]](https://leetcode.com/problems/3sum?envType=problem-list-v2&envId=atbyk0mj) [[42. Trapping Rain Water]](https://leetcode.com/problems/trapping-rain-water?envType=problem-list-v2&envId=atbyk0mj) [[56. Merge Intervals]](https://leetcode.com/problems/merge-intervals?envType=problem-list-v2&envId=atbyk0mj) [[57. Insert Interval]](https://leetcode.com/problems/insert-interval?envType=problem-list-v2&envId=atbyk0mj) [[46. Permutations]](https://leetcode.com/problems/permutations?envType=problem-list-v2&envId=atbyk0mj) [[53. Maximum Subarray]](https://leetcode.com/problems/maximum-subarray?envType=problem-list-v2&envId=atbyk0mj) [[73. Set Matrix Zeroes]](https://leetcode.com/problems/set-matrix-zeroes?envType=problem-list-v2&envId=atbyk0mj) [[284. Peeking Iterator]](https://leetcode.com/problems/peeking-iterator?envType=problem-list-v2&envId=atbyk0mj)\n\n####  [Interview Questions and Answers Repository](https://github.com/RooTinfinite/Interview-questions-and-answers)\n\n\n\n\n![image.png](https://assets.leetcode.com/users/images/9dc1b265-b175-4bf4-bc6c-4b188cb79220_1728176037.4402142.png)\n\n\n\n![image.png](https://assets.leetcode.com/users/images/25e711ac-aff3-4d9e-b1ad-fc79637d07a4_1731660426.9422147.png)\n\n
['Swift', 'C++', 'Java', 'Go', 'TypeScript', 'Python3', 'Rust', 'Ruby', 'JavaScript', 'C#']