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 in 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.
Java
class LeadersInArray { /*Java 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 System.out.print(arr[i] + " "); } } /* Driver program to test above functions */ public static void main(String[] args) { LeadersInArray lead = new LeadersInArray(); int arr[] = new int[]{16, 17, 4, 3, 5, 2}; int n = arr.length; lead.printLeaders(arr, n); }} |
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:
Java
class LeadersInArray { /* Java 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 */ System.out.print(max_from_right + " "); for (int i = size-2; i >= 0; i--) { if (max_from_right < arr[i]) { max_from_right = arr[i]; System.out.print(max_from_right + " "); } } } /* Driver program to test above functions */ public static void main(String[] args) { LeadersInArray lead = new LeadersInArray(); int arr[] = new int[]{16, 17, 4, 3, 5, 2}; int n = arr.length; lead.printLeaders(arr, n); }} |
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!
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