Maximum Index || | Problem of the Day || gfg potd

In this video, we will learn to solve the Maximum Index problem of geeks for geeks.

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[rapidcoders]

int maxIndexDiff(int arr[], int n) {
// code here
vector<int> left_min(n, INT_MAX);
vector<int> right_max(n, INT_MIN);
int ans = 0;
left_min[0] = arr[0];
right_max[n-1] = arr[n-1];
for(int i=1;i<n;++i){
left_min[i] = min(arr[i], left_min[i-1]);
}
for(int i = n-2;i>=0;--i){
right_max[i] = max(arr[i], right_max[i+1]);
}
// for(auto i:left_min) cout<<i<<" ";
// cout<<endl;
// for(auto i:right_max) cout<<i<<" ";
// cout<<endl;

int lptr = 0, rptr = 0;
while(lptr < n && rptr < n){
if(left_min[lptr] <= right_max[rptr]){
// cout<<lptr<<" "<<rptr<<endl;
ans = max(ans, rptr-lptr);
++rptr;
}else{
++lptr;
}
}
return ans;
}

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[TodayContestSolution]

any one explain from any eg. if we use (left_Min[left_S] <right_Max[right_S])in place of (left_Min[left_S] <= right_Max[right_S])

[Priyanshukumar-ylnh]

its not a sliding window brother, its actually a 2 pointer approach