#114

[@shameem.pullaratt]

At 8:25, Is the rightSum = totalSum - a[i] - leftSum ?

[@PlacementsReady]

Yes please go through full video u will get it

[@BhavanaBharatiSingh]

@@PlacementsReady but it wont work it we have first position as equilibrium point for e.g: 1,0 so we need to use rightsum = totalsum - a[i], leftsum += a[i] and consdition to return is if leftsum == rightsum-a[i]

[@riyasingh6864]

Very good explaination

[@PlacementsReady]

Thanks Riya. Glad that you liked it.

[@farhanmuhamed392]

Two pointer method is more optimised, only takes O(n) time instead of O(2n)

[@cosmosXverse]

can you explain little bit...about two pointer method

[@PlacementsReady]

@farhanmuhamed392 how are you solving this question using two pointer approach?

[@farhanmuhamed392]

@@PlacementsReady int equilibriumPoint(long long a[], int n) { int left=0, right=n-1; long long sumLeft=a[left], sumRight=a[right]; while(left != right){ if(sumLeft > sumRight){ right--; sumRight = sumRight + a[right]; } else{ left++; sumLeft = sumLeft + a[left]; } } if( sumLeft == sumRight) return left+1; else return -1; }

[@farhanmuhamed392]

@@cosmosXverse Declare 4 variables left, right, leftSum, rightSum left and right are pointers , and leftSum and rightSum keep track sum of value at current pointers int equilibriumPoint(long long a[], int n) { int left=0, right=n-1; long long sumLeft=a[left], sumRight=a[right]; while(left != right){ if(sumLeft > sumRight){ right--; sumRight = sumRight + a[right]; } else{ left++; sumLeft = sumLeft + a[left]; } } if( sumLeft == sumRight) return left+1; else return -1; }

[@pranavmittal2976]

@@farhanmuhamed392 bro it is failing for the test case 1 2 0 3 it expected output is 3 but it is printing -1