Find Leaders in an Array Optimal Right to Left Scan Solution

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ProblemExamplesVisualizationSolutionAlgorithmCodeTimeSpace

Problem Statement❯

Given an array of integers, your task is to find all the leader elements in the array.

  • A leader is an element that is greater than all the elements to its right in the array.
  • The rightmost element is always a leader because there are no elements after it.

You need to return a list of all such leaders in left-to-right order as they appear in the array.

Examples❯

Input Array Leaders Description
[16, 17, 4, 3, 5, 2] [17, 5, 2] 17 > [4,3,5,2], 5 > 2, and 2 is last so it's a leader
[1, 2, 3, 4, 5] [5] Only 5 is greater than all elements to its right (none)
[5, 4, 3, 2, 1] [5, 4, 3, 2, 1] Each element is greater than all elements to its right
[7] [7] Single element is always a leader
[] [] Empty array, so no leaders
[10, 10, 10] [10] Only the last occurrence is considered a leader
[0, -1, -2, -3] [0, -1, -2, -3] All elements are leaders in descending order

Visualization Player

Solution❯

To solve the problem of finding leaders in an array, we need to understand what makes an element a leader. A leader is any number in the array that is strictly greater than all the numbers to its right.

We always start checking from the end of the array. This is because the last element has no elements after it, so it is always a leader by definition.

Different Cases Explained

  • Case 1: Multiple leaders exist – For example, in [16, 17, 4, 3, 5, 2], we start from the end:
    • 2 is a leader (no elements after it)
    • 5 > 2, so 5 is a leader
    • 3 < 5, so it’s not
    • 4 < 5, so skip
    • 17 > 5, so it's also a leader
    Final leaders in left-to-right order: [17, 5, 2]
  • Case 2: Increasing array – In [1, 2, 3, 4, 5], only 5 is a leader because no number is greater than it to the right.
  • Case 3: Decreasing array – In [5, 4, 3, 2, 1], each number is greater than the rest of the elements to its right, so every number is a leader.
  • Case 4: Equal elements – In [10, 10, 10], only the last 10 is a leader, because the previous ones are not greater than the next.
  • Case 5: Single element – Any array with just one element like [7] will always have that element as a leader.
  • Case 6: Empty array – No elements mean no leaders. Return an empty list.

By starting from the right and keeping track of the maximum element seen so far, we can easily identify all leader elements. Each time we find a new maximum (while scanning from right to left), we record it as a leader.

In the end, we reverse the list of leaders (since we found them from right to left) to return them in left-to-right order as required.

This approach is optimal with a time complexity of O(n) because we only scan the array once.

Algorithm Steps❯

  1. Given an array arr.
  2. Initialize an empty list leaders to store leader elements.
  3. Start from the last element: initialize max_from_right = arr[n-1].
  4. Add max_from_right to the leaders list.
  5. Traverse the array from right to left:
  6. → If current element is greater than max_from_right, update max_from_right and add it to leaders.
  7. After the loop, reverse the leaders list to maintain left-to-right order.

Code

Python
JavaScript
Java
C++
C
def find_leaders(arr):
    n = len(arr)
    leaders = []
    max_from_right = arr[-1]
    leaders.append(max_from_right)
    for i in range(n - 2, -1, -1):
        if arr[i] > max_from_right:
            max_from_right = arr[i]
            leaders.append(max_from_right)
    leaders.reverse()
    return leaders

# Sample Input
arr = [16, 17, 4, 3, 5, 2]
print("Leaders:", find_leaders(arr))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)The array is scanned from right to left once, and each element is checked only once, making the operation linear.
Average CaseO(n)Regardless of input distribution, the array is always scanned once from end to start to compare with the running maximum.
Worst CaseO(n)In the worst case (e.g., strictly decreasing array), every element becomes a leader, but the time taken remains linear as there is still only one pass.

Space Complexity

O(n)

Explanation: The worst-case space is O(n) if all elements are leaders and stored in the list. Additional space is used only for the leaders list and a few variables.