18. Leaders in an array

The problem can be found at the following link: Question Link

My Approach

To find the leaders in an array, my approach follows these steps:

• Traverse the array from right to left.

• Maintain a variable to keep track of the maximum element encountered so far.

• If the current element is greater than or equal to the maximum element encountered so far, add it to the output vector of leaders.

Explanation with Example

Let's consider an example to illustrate the approach:

Input array: [16, 17, 4, 3, 5, 2] Output leaders: [17, 5, 2]

Starting from the right end of the array, we encounter 2 which is the maximum among the remaining elements. So, 2 is a leader. Next, 5 is the maximum among the remaining elements, so it's also a leader. Finally, 17 is the maximum among the remaining elements, making it a leader as well.

Time and Auxiliary Space Complexity

• Time Complexity: The algorithm iterates through the array once in reverse order, which takes O(n) time, where n is the size of the array.

• Auxiliary Space Complexity: The extra space used is for the output vector, which stores the leaders. This takes O(k) space, where k is the number of leaders in the array.

Code (C++)

class Solution {
public:
    vector<int> leaders(int a[], int n) {
        vector<int> out;
        int nax = -1;

        for (int i = n - 1; i >= 0; --i) {
            if (nax <= a[i]) {
                nax = a[i];
                out.push_back(a[i]);
            }
        }

        reverse(out.begin(), out.end());
        return out;
    }
};

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