Can you tell me where is mistake?

#1


Can you tell me some other efficient approach.

#2

@raj.verma5454,
The problem breaks down to, find the last element that is greater than current element, so this can be done by just a single stack.

Maintain a stack that stores indexes of elements:

Case 1: a[i] <= a[s.top()]

  • Simply push the current index on top of stack, i.e s.push(i);

Case 2: a[i] > a[s.top()]

  • Now all the previous elements that are smaller than a[i] are useless for future, so pop all of them, i.e, pop till a[s.top()] <a[I], and then s.push(i);
#3

okk , but can you tell me for which test case my code failed?

#4

I am stuck. please help me

#5

@raj.verma5454,
Check your code against this test case:

  • Input :
    6
    1 2 3 1 2 4

  • Output:
    1 2 3 1 2 6

#6

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