Equilibrium Element in Array in o(n) time complexity ( Interview Question for FAANG companies)

Find equilibrium element in array in O(n) time complexity and o(n) space complexity. This is very important question for product based compnies and the FAANG compnies.

[yaamraj]

i have an issue with this question for a long time
now it is solved
thanks sir

[hariprasadcr]

best simple and strightforward solution

[rohithbharathi]

thank you for the detailed explanation

[hardikachordia]

Hi Can you please provide optimize solution if we have internal table of positive and negative integers with at least one positive integer, get the maximum possible positive sum

[devanshprakash]

Can u please upload code videos on bfs, dfs, prims algo, kruskal algo etc etc also ☺️

[Ahadi]

Please explain Cage- amalgamation graph, how we cane find it?
thanks

[gamingsparrow]

Sir please take class about hypergraph

[saranvasanth]

if the matrix is not balanced then how can I print it

[KNT_]

Here also we are using two loops. So I think this also n*n time complexity algo.

[viveksharma]

Here time complexity is O(2n) or O(2^n) which is exponential hence for larger input size it is not approx equals to O(n)

[amanahmed]

sir what if there is no equilibirum element

[banutejagoud]

It is wrong at the step rightsum=sum=0; becuase you have updated the sum in the above steps by using for loop . Then how the rightsum will be equal to zero at first?

[vivekmishra]

U must have explained the logic/mathematics behind the for loop

[krishnashejul]

int equilibriumPoint(long long a[], int n) {

// Your code here
int left[n];
left[0]=0;
for(int i=1;i<n;i++){
left[i]=a[i-1]+left[i-1];
}
int right[n];
right[n-1]=0;
for(int i=n-2;i>=0;i--){
right[i]=right[i+1]+a[i+1];
}
int result;
for(int i=0;i<n;i++){
if(left[i]==right[i]){
result=i;
}

}
if(result<n){
return result+1;
}
else
return -1;
}
Is this correct way to above problem ??