Detailed and well explained C++ Solution ✔️✔️✔️
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n1. Create a 2D vector v of size n by 2, where v[i][0] stores the value of
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. Create a 2D vector v of size n by 2, where v[i][0] stores the value of nums2[i] and v[i][1] stores the value of nums1[i].\n\n2. Sort the 2D vector v based on the values in nums2 in non-decreasing order. Sorting helps in selecting the optimal elements for the operations.\n\n3. Initialize a vector dp of size n+1 to keep track of the maximum values that can be obtained for each position in nums2 by considering operations involving elements from both nums1 and nums2.\n\n4. The first loop iterates from j = 1 to j = n. Within this loop, there is another loop that iterates over i in reverse order, starting from i = j and going down to i = 1.\n\n5. In the nested loop, calculate the maximum value that can be obtained for the operation at position j considering i elements from nums2 and elements from nums1. Update dp[i] with the maximum of two values:\n\n6. The current value of dp[i].\nThe sum of dp[i-1] + i * v[j-1][0] + v[j-1][1], representing the maximum value we can save by selecting i elements from nums2 and performing the operation with elements at positions i and j.\nAfter the first loop, the dp vector will contain the maximum values that can be obtained for each position in nums2 by considering operations involving elements from both nums1 and nums2.\n\n7. Calculate the sum of all elements in nums1 and store it in the integer variable suma. Calculate the sum of all elements in nums2 and store it in the integer variable sumb.\n\n8. The second loop iterates from i = 0 to i = n. Within this loop, check if the condition sumb * i + suma - dp[i] <= x is satisfied.\n\n9. If the condition is satisfied, it means that selecting i elements from nums2 and performing operations with elements from both nums1 and nums2 will not exceed the constraint x. In this case, the function returns the value of i, which represents the minimum time needed to meet the constraint.\n\n10. If the condition is not satisfied for any value of i within the loop, it means that there is no valid combination of elements that satisfies the constraint x. In this case, the function returns -1 to indicate that the constraint cannot be met.\n\n# Complexity\n- Time complexity:\nO(n^2)\n\n- Space complexity:\nO(n)\n\n# Code\n```\nclass Solution {\npublic:\n    int minimumTime(vector<int>& nums1, vector<int>& nums2, int x) {\n        int n = size(nums1);\n        vector<int> dp(n+1);\n        vector<vector<int>> v(n, vector<int>(2));\n        for(int i = 0; i<n; ++i) v[i][0] = nums2[i], v[i][1] = nums1[i];\n        sort(v.begin(), v.end());\n        for(int j = 1; j<=n; ++j)\n        for(int i = j; i>=1; --i) dp[i] = max(dp[i], dp[i-1] + i*v[j-1][0] + v[j-1][1]);\n        int suma = accumulate(nums1.begin(), nums1.end(), 0);\n        int sumb = accumulate(nums2.begin(), nums2.end(), 0);\n        for(int i = 0; i<=n; ++i)\n        if(sumb*i + suma - dp[i] <= x) return i;\n        return -1;\n    }\n};\n```
['C++']