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Leetcode Problem Link: 724. Find Pivot Index

*Brute Force Solution:*

```
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;
}
};
```

*Efficient Solution:*

```
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;
}
};
```

*All suggestions are welcome. Please upvote if you like it. Thank you.*

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