724. Find Pivot Index [Leetcode][C++]

class Solution { public: int pivotIndex(vector<int>& nums) { // Brute Force Solution Time O(N^2) & Auxiliary Space O(1) // Calculate the sum of elements to the left & right // of each index of nums vector and compare them to check equivalence int len=nums.size(); if(len==1) return nums[0]; for(int i=0;i<len;i++){ // accumulate() takes O(N) time and for loop runs for n iterations. // So time taken by for loop is O(n*n)=O(N^2) int left=accumulate(nums.begin(),nums.begin()+i,0); int right=accumulate(nums.begin()+i+1,nums.end(),0); if(left==right) return i; } return -1; } }; 

class Solution { public: int pivotIndex(vector<int>& nums) { // Efficient Solution Time O(N) & Auxiliary Space O(1) // Create a sum array in which each element stores  // the sum of elements of nums vector up to that index. // Calculate left & right sum. For example- // nums[]=[1,7,3,6,5,6] // sum[]= [1,8,11,17,22,28] // left[3]=17-6=11 // right[3]=28-17=11 // return 3 int len=nums.size(); if(len==1) return 0; int sum[len]; sum[0]=nums[0]; for(int i=1;i<len;i++){ sum[i]=nums[i]+sum[i-1]; } for(int i=0;i<len;i++) { int left=sum[i]-nums[i]; int right=sum[len-1]-sum[i]; if(left==right) return i; } return -1; } };