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.
Javascript
<script> // Javascript Function to print leaders in an array function printLeaders( arr, size) { for (let i = 0; i < size; i++) { let j; for (j = i+1; j < size; j++) { if (arr[i] <=arr[j]) break ; } if (j == size) // the loop didn't break document.write(arr[i] + " " ); } } // driver code let arr = [ 16, 17, 4, 3, 5, 2 ]; let n = arr.length; // Function calling printLeaders(arr, n); </script> |
Output:
17 5 2
Time Complexity: O(n*n)
Auxiliary Space: O(1)
As constant extra space is used.
Method 2 (Scan from right)
Scan all the elements from right to left in an array and keep track of maximum till now. When maximum changes its value, print it.
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
Javascript
<script> /* JavaScript Function to print leaders in an array */ function printLeaders(arr,size) { let max_from_right = arr[size-1]; /* Rightmost element is always leader */ document.write(max_from_right + " " ); for (let i = size-2; i >= 0; i--) { if (max_from_right < arr[i]) { max_from_right = arr[i]; document.write(max_from_right + " " ); } } } /* Driver program to test above function*/ let arr = [16, 17, 4, 3, 5, 2]; let n = arr.length; printLeaders(arr, n); </script> |
Output:
2 5 17
Time Complexity: O(n)
Auxiliary Space: O(n)
The extra space is used to store the elements of max_from_right array.
Please refer complete article on Leaders in an array for more details!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!