Problem Overview & Intuition

Checking from left to right would require an O(N) search for each element (O(N²)). By traversing from right to left, we maintain maxFromRight in O(1) time per element.

Step-by-Step Approach

1. Initialize result = [] and maxFromRight = -Infinity.
2. Iterate i from arr.length - 1 down to 0.
3. If arr[i] >= maxFromRight, push arr[i] to result and update maxFromRight = arr[i].
4. Reverse result to maintain original left-to-right order, then return it.

Optimal Implementation (JavaScript)

function leaders(arr) {
  const result = [];
  let maxFromRight = -Infinity;
  for (let i = arr.length - 1; i >= 0; i--) {
    if (arr[i] >= maxFromRight) {
      result.push(arr[i]);
      maxFromRight = arr[i];
    }
  }
  return result.reverse();
}

Complexity Analysis