Find position of the leader element in given Array with given operations

Last Updated : 27 Jan, 2022

Given an array arr[]. The task is to find the position of the leader element in arr[]. The leader element is the one, which can remove all other elements in the array using the below operations.

If arr[i] > arr[i + 1], It removes the (i+1)th element and increment their value by 1 and decrease the size of array by 1.
If arr[i] > arr[i - 1], It removes the (i-1)th element and increment their value by 1 and decrease the size of array by 1.

Examples

Input: arr[] = { 5, 3, 4, 4, 5 }
Output: 3
Explanation: Following are the operations performed in array arr[]
A3 remove A2 and increment by 1 and array becomes { 5, 5, 4, 5 }
A2 remove A3 and increment by 1 and array becomes { 5, 6, 5 }
A2 remove A1 and increment by 1 and array becomes { 7, 5 }
A1 remove A2 and increment by 1 and array becomes { 8 }
Hence, The position of leader of array is 3.

Input: arr[] = { 4, 4, 3, 4, 4 }
Output: 2
Explanation: Following are the operations performed in array arr[]
A2 remove A3 and increment by 1 and array becomes { 4, 5, 4, 4 }
A2 remove A1 and increment by 1 and array becomes { 6, 4, 4 }
A1 remove A2 and increment by 1 and array becomes { 7, 4 }
A1 remove A2 and increment by 1 and array becomes { 8 }
Hence, The position of leader of array is 2.

Input: arr[] = { 1, 1, 1 }
Output: -1
Explanation: No leader is present in the array

Approach: This problem is implementation-based. Follow the steps below to solve the given problem.

Find the maximum and minimum elements of the array arr.
If the minimum and maximum elements are the same that means no leader element is present.
Traverse the array.
- Now, check if arr[i] is max element.
- Check arr[i-1] or arr[i+1] is smaller than max .
- So, that element is the leader element in the array.
Print the position of the leader element found.

Below is the implementation of the above approach.

C++

// C++ program for above approach #include <bits/stdc++.h> using namespace std;  // Function to find the leader element in array int LeaderElement(int arr[], int n) {      // Initialize two variable     int maxElement = INT_MIN,         minElement = INT_MAX;      // Traverse the array     for (int i = 0; i < n; i++) {         maxElement = max(maxElement, arr[i]);         minElement = min(minElement, arr[i]);     }      // Now if both are equal return -1     if (maxElement == minElement) {         return -1;     }     // Now traverse the array and     // check if adjacent element     // of max element because     // maxelement of array always     // leader But if more than 1     // leader element so, check     // the adjacent elements of that     int ans = -1;     for (int i = 0; i < n; i++) {         if (arr[i] == maxElement) {             if (i > 0                 and arr[i] > arr[i - 1]) {                 ans = i + 1;                 break;             }             if (i < n - 1                 and arr[i] > arr[i + 1]) {                 ans = i + 1;                 break;             }         }     }     return ans; }  // Driver Code int main() {     int arr[] = { 4, 4, 3, 4, 4 };      int N = 5;      // Function Call     int ans = LeaderElement(arr, N);     cout << ans;     return 0; }

Java

// Java program for the above approach import java.io.*; import java.lang.*; import java.util.*;  class GFG  {      // Function to find the leader element in array   static  int LeaderElement(int arr[], int n)   {      // Initialize two variable     int maxElement = Integer.MIN_VALUE;     int  minElement =  Integer.MAX_VALUE;      // Traverse the array     for (int i = 0; i < n; i++) {       maxElement = Math.max(maxElement, arr[i]);       minElement = Math.min(minElement, arr[i]);     }      // Now if both are equal return -1     if (maxElement == minElement) {       return -1;     }     // Now traverse the array and     // check if adjacent element     // of max element because     // maxelement of array always     // leader But if more than 1     // leader element so, check     // the adjacent elements of that     int ans = -1;     for (int i = 0; i < n; i++) {       if (arr[i] == maxElement) {         if (i > 0 && arr[i] > arr[i - 1]) {           ans = i + 1;           break;         }         if (i < n - 1 && arr[i] > arr[i + 1]) {           ans = i + 1;           break;         }       }     }     return ans;   }   public static void main (String[] args) {      int arr[] = { 4, 4, 3, 4, 4 };     int N = 5;      // Function Call     int ans = LeaderElement(arr, N);     System.out.print(ans);   } }  // This code is contributed by hrithikgarg03188

Python

# Python3 program for the above approach # import the module import sys  # Function to find the leader element in array def LeaderElement(arr, n):      # Initialize two variable     maxElement = -sys.maxsize - 1     minElement =  sys.maxint      # Traverse the array     for i in range(n):                  maxElement = max(maxElement, arr[i])         minElement = min(minElement, arr[i])           # Now if both are equal return -1     if (maxElement == minElement):         return -1          # Now traverse the array and     # check if adjacent element     # of max element because     # maxelement of array always     # leader But if more than 1     # leader element so, check     # the adjacent elements of that     ans = -1     for i in range(n):              if (arr[i] == maxElement):              if (i > 0 and arr[i] > arr[i - 1]):                 ans = i + 1                 break                          if (i < n - 1 and arr[i] > arr[i + 1]):                  ans = i + 1                 break               return ans   # Driver Code  # Given Input arr = [ 4, 4, 3, 4, 4 ] N = len(arr)  # Function Call ans = LeaderElement(arr, N) print(ans)     # This code is contributed by hrithikgarg03188.

// C# program for the above approach using System; class GFG  {      // Function to find the leader element in array   static  int LeaderElement(int []arr, int n)   {      // Initialize two variable     int maxElement = Int32.MinValue;     int minElement = Int32.MaxValue;      // Traverse the array     for (int i = 0; i < n; i++) {       maxElement = Math.Max(maxElement, arr[i]);       minElement = Math.Min(minElement, arr[i]);     }      // Now if both are equal return -1     if (maxElement == minElement) {       return -1;     }          // Now traverse the array and     // check if adjacent element     // of max element because     // maxelement of array always     // leader But if more than 1     // leader element so, check     // the adjacent elements of that     int ans = -1;     for (int i = 0; i < n; i++) {       if (arr[i] == maxElement) {         if (i > 0 && arr[i] > arr[i - 1]) {           ans = i + 1;           break;         }         if (i < n - 1 && arr[i] > arr[i + 1]) {           ans = i + 1;           break;         }       }     }     return ans;   }   public static void Main () {      int []arr = { 4, 4, 3, 4, 4 };     int N = 5;      // Function Call     int ans = LeaderElement(arr, N);     Console.Write(ans);   } }  // This code is contributed by Samim Hossain Mondal.

JavaScript

<script> // Javascript program for above approach   // Function to find the leader element in array function LeaderElement(arr, n) {      // Initialize two variable     let maxElement = Number.MIN_SAFE_INTEGER,      minElement = Number.MIN_SAFE_INTEGER;      // Traverse the array     for (let i = 0; i < n; i++) {         maxElement = Math.max(maxElement, arr[i]);         minElement = Math.min(minElement, arr[i]);     }      // Now if both are equal return -1     if (maxElement == minElement) {         return -1;     }     // Now traverse the array and     // check if adjacent element     // of max element because     // maxelement of array always     // leader But if more than 1     // leader element so, check     // the adjacent elements of that     let ans = -1;     for (let i = 0; i < n; i++) {         if (arr[i] == maxElement) {             if (i > 0 && arr[i] > arr[i - 1]) {                 ans = i + 1;                 break;             }             if (i < n - 1 && arr[i] > arr[i + 1]) {                 ans = i + 1;                 break;             }         }     }     return ans; }  // Driver Code let arr = [4, 4, 3, 4, 4];  let N = 5;  // Function Call let ans = LeaderElement(arr, N); document.write(ans)  // This code is contributed by gfgking. </script>