Leaders in an array

Write a program to print all the LEADERS in the array. An element is leader if it is greater than all the elements to its right side. And the rightmost element is always a leader. For example int the array {16, 17, 4, 3, 5, 2}, leaders are 17, 5 and 2.

Let the input array be arr[] and size of the array be size.

Method 1 (Simple)
Use two loops. The outer loop runs from 0 to size – 1 and one by one picks all elements from left to right. The inner loop compares the picked element to all the elements to its right side. If the picked element is greater than all the elements to its right side, then the picked element is the leader.

• C++
• Python

#include<iostream> using namespace std;   /*C++ Function to print leaders in an array */ void printLeaders(int arr[], int size) {     for (int i = 0; i < size; i++)     {         int j;         for (j = i+1; j < size; j++)         {             if (arr[i] <= arr[j])                 break;         }            if (j == size) // the loop didn't break             cout << arr[i] << " ";   } }   /* Driver program to test above function */ int main() {     int arr[] = {16, 17, 4, 3, 5, 2};     int n = sizeof(arr)/sizeof(arr[0]);     printLeaders(arr, n);     return 0; }

Output:

17 5 2

Time Complexity: O(n*n)

Method 2 (Scan from right)
Scan all the elements from right to left in array and keep track of maximum till now. When maximum changes it’s value, print it.

• C++
• Python

#include <iostream> using namespace std;   /* C++ Function to print leaders in an array */ void printLeaders(int arr[], int size) {     int max_from_right =  arr[size-1];       /* Rightmost element is always leader */     cout << max_from_right << " ";           for (int i = size-2; i >= 0; i--)     {         if (max_from_right < arr[i])         {                       max_from_right = arr[i];             cout << max_from_right << " ";         }     }    }   /* Driver program to test above function*/ int main() {     int arr[] = {16, 17, 4, 3, 5, 2};     int n = sizeof(arr)/sizeof(arr[0]);     printLeaders(arr, n);     return 0; }

Output

2 5 17

Time Complexity: O(n)

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