Array Leaders

Easy Arrays & Hashing Right to Left Traversal Original on GeeksforGeeks

Array Leaders is a easy arrays & hashing problem solved with the right to left traversal pattern. The best approach, optimal (right-to-left max), runs in O(n) time and O(1) space. Below are 2 approaches in Java, from brute force up.

Problem

An element is a leader if it is greater than or equal to every element to its right. The rightmost element is always a leader. Return all leaders in the order they appear in the array.

Examples

Example 1

Input: arr = [10, 4, 2, 4, 1]
Output: [10, 4, 4, 1]
Why: 10 beats everything after it; the second 4 ties nothing larger after it; 1 is last.

Example 2

Input: arr = [5, 7, 3]
Output: [7, 3]

Constraints

1 <= arr.length <= 10^6
0 <= arr[i] <= 10^6
An element is a leader if it is greater than or equal to every element to its right.

Animated walkthrough

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Brute force	`O(n²)`	`O(1)`
Optimal (right-to-left max)	`O(n)`	`O(1)`

1Brute force

TimeO(n²)

SpaceO(1)Besides the output list.

For each element, scan everything to its right and check that nothing is larger.

For each i, check arr[j] <= arr[i] for all j > i.
Collect i if the check passes.

Java

class Solution {
    static ArrayList<Integer> leaders(int[] arr) {
        ArrayList<Integer> out = new ArrayList<>();
        for (int i = 0; i < arr.length; i++) {
            boolean leader = true;
            for (int j = i + 1; j < arr.length; j++)
                if (arr[j] > arr[i]) { leader = false; break; }
            if (leader) out.add(arr[i]);
        }
        return out;
    }
}

2Optimal (right-to-left max)

TimeO(n)

SpaceO(1)Besides the output list.

Traverse from the right with the maximum seen so far. An element is a leader when it is at least that maximum. Collect leaders in reverse, then reverse the list to restore the original order.

max = -infinity.
For i from n - 1 to 0: if arr[i] >= max, add it and set max = arr[i].
Reverse the collected list.

Java

class Solution {
    static ArrayList<Integer> leaders(int[] arr) {
        ArrayList<Integer> out = new ArrayList<>();
        int max = Integer.MIN_VALUE;
        for (int i = arr.length - 1; i >= 0; i--) {
            if (arr[i] >= max) {
                out.add(arr[i]);
                max = arr[i];
            }
        }
        Collections.reverse(out);
        return out;
    }
}

Edge cases to test

The last element is always a leader
Equal values (>= counts as a leader)
A strictly increasing array has only one leader

Hints

Hint 1

Walk from the right while tracking the maximum seen so far.

FAQ

What is the best time complexity for Array Leaders?

Optimal (right-to-left max) runs in O(n) time and O(1) extra space.

Which pattern does Array Leaders use?

It is a arrays & hashing problem that uses the right to left traversal pattern. Other problems with the same pattern: Product of Array Except Self.

Is there a brute force solution for Array Leaders?

Yes. Brute force takes O(n²) time and O(1) space. For each element, scan everything to its right and check that nothing is larger.

Which edge cases should I test for Array Leaders?

The last element is always a leader; Equal values (= counts as a leader); A strictly increasing array has only one leader.