Write a program to print all the LEADERS in the array. An element is a leader if it is greater than all the elements to its right side. And the rightmost element is always a leader.
For example:
Input: arr[] = {16, 17, 4, 3, 5, 2},
Output: 17, 5, 2Input: arr[] = {1, 2, 3, 4, 5, 2},
Output: 5, 2
Naive Approach: The problem can be solved based on the idea mentioned below:
Use two loops. The outer loop runs from 0 to size – 1 and one by one pick all elements from left to right. The inner loop compares the picked element to all the elements on its right side. If the picked element is greater than all the elements to its right side, then the picked element is the leader.
Follow the below steps to implement the idea:
- We run a loop from the first index to the 2nd last index.
- And for each index, we run another loop from the next index to the last index.
- If all the values to the right of that index are smaller than the index, we simply add the value in our answer data structure.
Below is the implementation of the above approach.
C
/*C Function to print leaders in an array */#include <stdio.h>void printLeaders(int arr[], int size){ int i, j; for (i = 0; i < size; i++) { for (j = i + 1; j < size; j++) { if (arr[i] <= arr[j]) break; } if (j == size) // the loop didn't break printf("%d ", 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;} |
C++
#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;} |
Java
import java.io.*;public 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); }} |
Python3
# Python Function to print leaders in array def printLeaders(arr,size): for i in range(0, size): for j in range(i+1, size): if arr[i]<=arr[j]: break if j == size-1: # If loop didn't break print (arr[i],end=' ') # Driver function arr=[16, 17, 4, 3, 5, 2] printLeaders(arr, len(arr)) # This code is contributed by _Devesh Agrawal__ |
C#
// C# program to print// leaders in arrayusing System;class GFG { 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; } // the loop didn't break if (j == size) Console.Write(arr[i] + " "); } } // Driver Code public static void Main() { GFG lead = new GFG(); int []arr = new int[]{16, 17, 4, 3, 5, 2}; int n = arr.Length; lead.printLeaders(arr, n); }}// This code is contributed by// Akanksha Rai(Abby_akku) |
PHP
<?php// PHP Function to print// leaders in an array function printLeaders($arr, $size){ for ($i = 0; $i < $size; $i++) { for ($j = $i + 1; $j < $size; $j++) { if ($arr[$i] <=$arr[$j]) break; } // the loop didn't break if ($j == $size) echo($arr[$i] . " "); }}// Driver Code$arr = array(16, 17, 4, 3, 5, 2);$n = sizeof($arr);printLeaders($arr, $n); // This code is contributed // by Shivi_Aggarwal?> |
Javascript
<script>// Javascript Function to print leaders in an arrayfunction 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)
Find Leader by finding suffix maximum:
The idea is to scan all the elements from right to left in an array and keep track of the maximum till now. When the maximum changes its value, print it.
Follow the below illustration for a better understanding
Illustration:
Let the array be arr[] = {16, 17, 4, 3, 5, 2}
- arr[] = {16, 17, 4, 3, 5, 2} , max_from_right = 2 , ans[] = { 2 }
- arr[] = {16, 17, 4, 3, 5, 2} , max_from_right = 5 , ans[] = { 2, 5 }
- arr[] = {16, 17, 4, 3, 5, 2} , max_from_right = 5 , ans[] = { 2, 5 }
- arr[] = {16, 17, 4, 3, 5, 2} , max_from_right = 5 , ans[] = { 2, 5 }
- arr[] = {16, 17, 4, 3, 5, 2} , max_from_right = 17 , ans[] = { 2, 5, 17 }
- arr[] = {16, 17, 4, 3, 5, 2} , max_from_right = 17 , ans[] = { 2, 5, 17 }
Follow the steps mentioned below to implement the idea:
- We start from the last index position. The last position is always a leader, as there are no elements towards its right.
- And then we iterate on the array till we reach index position = 0.
- Each time we keep a check on the maximum value
- Every time we encounter a maximum value than the previous maximum value encountered, we either print or store the value as it is the leader
Below is the implementation of the above approach.
C
#include <stdio.h>/* 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 */ printf("%d ", max_from_right); for (int i = size - 2; i >= 0; i--) { if (max_from_right < arr[i]) { max_from_right = arr[i]; printf("%d ", 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;} |
C++
#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;} |
Java
import java.io.*;public 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); }} |
Python3
# Python function to print leaders in arraydef printLeaders(arr, size): max_from_right = arr[size-1] print (max_from_right,end=' ') for i in range( size-2, -1, -1): if max_from_right < arr[i]: print (arr[i],end=' ') max_from_right = arr[i] # Driver functionarr = [16, 17, 4, 3, 5, 2]printLeaders(arr, len(arr))# This code contributed by _Devesh Agrawal__ |
C#
// C# program to find Leaders in an arrayusing System;class LeadersInArray { // 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 Console.Write(max_from_right +" "); for (int i = size - 2; i >= 0; i--) { if (max_from_right < arr[i]) { max_from_right = arr[i]; Console.Write(max_from_right +" "); } } } // Driver Code 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); }}// This code is contributed // by Akanksha Rai(Abby_akku) |
PHP
<?php// PHP Function to print// leaders in an array function printLeaders(&$arr, $size){ $max_from_right = $arr[$size - 1]; // Rightmost element // is always leader echo($max_from_right); echo(" "); for ($i = $size - 2; $i >= 0; $i--) { if ($max_from_right < $arr[$i]) { $max_from_right = $arr[$i]; echo($max_from_right); echo(" "); } } }// Driver Code$arr = array(16, 17, 4, 3, 5, 2);$n = sizeof($arr);printLeaders($arr, $n);// This code is contributed // by Shivi_Aggarwal?> |
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(1)
Find leaders and print them in the same order as they are:
In the previous method, we get time linear complexity, but the output we get is not in the same order as the elements that appear in our input array, so to get out output in the same order as in the input array, we can use stack data structure.
Illustration:
Let the array be arr[] = {16, 17, 4, 3, 5, 2}, we will store the ans in a stack to print in the same order.
- arr[] = {16, 17, 4, 3, 5, 2} , max_from_right = 2 , ans[] = { 2 }
- arr[] = {16, 17, 4, 3, 5, 2} , max_from_right = 5 , ans[] = { 5, 2 }
- arr[] = {16, 17, 4, 3, 5, 2} , max_from_right = 5 , ans[] = { 5, 2 }
- arr[] = {16, 17, 4, 3, 5, 2} , max_from_right = 5 , ans[] = { 5, 2 }
- arr[] = {16, 17, 4, 3, 5, 2} , max_from_right = 17 , ans[] = { 17, 5, 2 }
- arr[] = {16, 17, 4, 3, 5, 2} , max_from_right = 17 , ans[] = { 17, 5, 2 }
Follow the below steps to implement the idea:
- We start from the last index position. The last position is always a leader, as there are no elements towards its right.
- And then we iterate on the array till we reach index position = 0.
- Each time we keep a check on the maximum value
- Every time we encounter a maximum value than the previous maximum value encountered, we will store the value in the stack as it is the leader
- We will iterate on the stack and print the values
Below is the implementation of the above approach.
C
#include <stdio.h>#include <stdlib.h>/* C Function to print leaders in an array */void printLeaders(int arr[], int size){ /* create stack to store leaders*/ int* sk = (int*)malloc(size * sizeof(int)); int top = -1; sk[++top] = arr[size - 1]; for (int i = size - 2; i >= 0; i--) { if (arr[i] >= sk[top]) { sk[++top] = arr[i]; } } /* print stack elements*/ /* run loop till stack is not empty*/ while (top != -1) { printf("%d ", sk[top--]); } free(sk);}/* 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;} |
C++
#include <bits/stdc++.h>using namespace std; /* C++ Function to print leaders in an array */void printLeaders(int arr[], int size){ /* create stack to store leaders*/ stack<int> sk; sk.push(arr[size-1]); for (int i = size-2; i >= 0; i--) { if(arr[i] >= sk.top()) { sk.push(arr[i]); } } /* print stack elements*/ /* run loop till stack is not empty*/ while(!sk.empty()){ cout<<sk.top()<<" "; sk.pop(); }} /* 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;} |
Java
// JAVA code for the above approachimport java.io.*;import java.util.*;class LeadersInArray { /* Java Function to print leaders in an array */ void printLeaders(int arr[], int size) { /* create stack to store leaders*/ Stack<Integer> stack = new Stack<Integer>(); stack.push(arr[size - 1]); for (int i = size - 2; i >= 0; i--) { if (arr[i] >= stack.peek()) { stack.push(arr[i]); } } /* print stack elements*/ /* run loop till stack is not empty*/ while (!stack.empty()) { System.out.print(stack.pop() + " "); } } /* Driver program to test above function*/ 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); }} |
Python3
# Python Function to print leaders in an arraydef printLaders(arr, size): # create stack to store leaders sk = [] sk.append(arr[size - 1]) for i in range(size - 2, -1, -1): if(arr[i] >= sk[len(sk) - 1]): sk.append(arr[i]) # print stack elements # run loop till stack is not empty while(len(sk) != 0): print(sk[len(sk)-1],end = ' ') sk.pop()# Driver program to test above functionif __name__ == "__main__": arr = [16,17,4,3,5,2] n = len(arr) printLaders(arr,n) # This code is contributed by ajaymakvana |
C#
// C# code for the above approachusing System;using System.Collections;public class GFG { // C# Function to print leaders in an array public static void printLeaders(int[] arr, int size) { // create stack to store leaders Stack stack = new Stack(); stack.Push(arr[size - 1]); for (int i = size - 2; i >= 0; i--) { if (arr[i] >= Convert.ToInt32(stack.Peek())) { stack.Push(arr[i]); } } // print stack elements // run loop till stack is not empty while (stack.Count > 0) { Console.Write(stack.Pop() + " "); } } // Driver program to test above function public static void Main(string[] args) { int[] arr = { 16, 17, 4, 3, 5, 2 }; int n = arr.Length; printLeaders(arr, n); }}// This code is contributed by ajaymakavana. |
Javascript
<script>function printLeaders(arr, size) { /* create stack to store leaders*/ let stack = []; stack.push(arr[size - 1]); for (let i = size - 2; i >= 0; i--) { let temp = stack.pop(); stack.push(temp); if (arr[i] >= temp) { stack.push(arr[i]); } } /* print stack elements*/ /* run loop till stack is not empty*/ while (stack.length>0) { console.log(stack.pop() + " "); } } let arr = [ 16, 17, 4, 3, 5, 2 ]; let n = arr.length; printLeaders(arr, n); </script> |
Output
17 5 2
Time complexity: O(n)
Auxiliary space: O(n)
Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.
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