easiest solution of c++ || using loops ||
Intuition\n\nCertainly! The intuition for the provided code is as follows:\n\nIntuition:\n\nThe problem requires us to determine if there exist two subarrays of
# Intuition\n\nCertainly! The intuition for the provided code is as follows:\n\nIntuition:\n\nThe problem requires us to determine if there exist two subarrays of consecutive elements in the given nums array such that the sum of the elements in the first subarray is equal to the sum of the elements in the second subarray.\n\nHere\'s the intuition behind the code:\n\nFirst, we calculate the sum of adjacent elements in the nums array and store these sums in the weights vector. This is done by iterating through nums from index 0 to n-2 and calculating sum = nums[i] + nums[i+1] for each pair of adjacent elements. These sums represent the possible sums of two consecutive elements.\n\nNext, we sort the weights vector in ascending order. Sorting is necessary because it helps us identify if there are consecutive equal sums, which would satisfy the problem\'s condition.\n\nAfter sorting, we iterate through the sorted weights vector and check if there are any two consecutive elements that are equal. If we find such elements, it means we\'ve found two subarrays with equal sums, and we return true. Otherwise, we return false at the end of the function.\n\nThe core idea is to transform the problem of finding two subarrays with equal sums into finding consecutive equal elements in the sorted list of possible sums. If we find such elements, we can conclude that there are indeed two subarrays with equal sums in the original array.\n\n# Approach\nThe problem asks us to determine whether there exist two subarrays of consecutive elements in the given nums array such that the sum of elements in the first subarray is equal to the sum of elements in the second subarray. We can solve this problem using the following approach:\n\nCreate a vector called weights to store the sums of adjacent elements in the nums array.\n\nIterate through the nums array from the first element to the second-to-last element. For each pair of adjacent elements at indices i and i+1, calculate their sum (sum = nums[i] + nums[i+1]) and add it to the weights vector. This step creates a list of possible sums of two consecutive elements in the original array.\n\nSort the weights vector in ascending order. Sorting is essential to identify consecutive equal sums efficiently.\n\nIterate through the sorted weights vector. For each element at index i, compare it with the next element at index i+1. If these two elements are equal, it implies that there are two subarrays in the original nums array with equal sums. In this case, return true.\n\nIf no consecutive equal sums are found while iterating through the sorted weights vector, return false at the end of the function.\n\nThe key insight here is that sorting the sums simplifies the problem because it allows us to identify consecutive equal sums efficiently. If we find consecutive equal sums in the sorted list, it implies that there are two subarrays in the original array with the same sum, satisfying the problem\'s condition.\n\nThis approach has a time complexity of O(n log n) due to the sorting step, where n is the length of the nums array, and a space complexity of O(n) to store the weights vector.\n# Complexity\n- Time complexity:\n O(n log n)\n- Space complexity:\n O(n) \n\n# Code\n```\n#include <vector>\n#include <algorithm> // for sort\nusing namespace std;\n\nclass Solution {\npublic:\n    bool findSubarrays(vector<int>& nums) {\n        vector<int> weights;\n        int n = nums.size();\n\n        for (int i = 0; i < n - 1; i++) {\n            int sum = nums[i] + nums[i + 1];\n            weights.push_back(sum);\n        }\n\n        sort(weights.begin(), weights.end());\n\n        for (int i = 0; i < weights.size() - 1; i++) {\n            int sum1 = weights[i];\n            int sum2 = weights[i + 1];\n            if (sum1 == sum2) {\n                return true;\n            }\n        }\n\n        return false;\n    }\n};\n\n```
['C++']