Given an array arr[] of size N. The task is to find maximum element among N – 1 elements other than arr[i] for each i from 1 to N.
Examples:
Input: arr[] = {2, 5, 6, 1, 3}
Output: 6 6 5 6 6
Input: arr[] = {1, 2, 3}
Output: 3 3 2
Approach: An efficient approach is to make prefix and suffix array of maximum elements and find maximum element among N – 1 elements other than arr[i].
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to find maximum element// among (N - 1) elements other than// a[i] for each i from 1 to Nint max_element(int a[], int n){ // To store prefix max element int pre[n]; pre[0] = a[0]; for (int i = 1; i < n; i++) pre[i] = max(pre[i - 1], a[i]); // To store suffix max element int suf[n]; suf[n - 1] = a[n - 1]; for (int i = n - 2; i >= 0; i--) suf[i] = max(suf[i + 1], a[i]); // Find the maximum element // in the array other than a[i] for (int i = 0; i < n; i++) { if (i == 0) cout << suf[i + 1] << " "; else if (i == n - 1) cout << pre[i - 1] << " "; else cout << max(pre[i - 1], suf[i + 1]) << " "; }}// Driver codeint main(){ int a[] = { 2, 5, 6, 1, 3 }; int n = sizeof(a) / sizeof(a[0]); max_element(a, n); return 0;} |
Java
// Java implementation of the approachclass GFG {// Function to find maximum element// among (N - 1) elements other than// a[i] for each i from 1 to Nstatic void max_element(int a[], int n){ // To store prefix max element int []pre = new int[n]; pre[0] = a[0]; for (int i = 1; i < n; i++) pre[i] = Math.max(pre[i - 1], a[i]); // To store suffix max element int []suf = new int[n]; suf[n - 1] = a[n - 1]; for (int i = n - 2; i >= 0; i--) suf[i] = Math.max(suf[i + 1], a[i]); // Find the maximum element // in the array other than a[i] for (int i = 0; i < n; i++) { if (i == 0) System.out.print(suf[i + 1] + " "); else if (i == n - 1) System.out.print(pre[i - 1] + " "); else System.out.print(Math.max(pre[i - 1], suf[i + 1]) + " "); }}// Driver codepublic static void main(String []args) { int a[] = { 2, 5, 6, 1, 3 }; int n = a.length; max_element(a, n);}}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach # Function to find maximum element # among (N - 1) elements other than # a[i] for each i from 1 to N def max_element(a, n) : # To store prefix max element pre = [0] * n; pre[0] = a[0]; for i in range(1, n) : pre[i] = max(pre[i - 1], a[i]); # To store suffix max element suf = [0] * n; suf[n - 1] = a[n - 1]; for i in range(n - 2, -1, -1) : suf[i] = max(suf[i + 1], a[i]); # Find the maximum element # in the array other than a[i] for i in range(n) : if (i == 0) : print(suf[i + 1], end = " "); elif (i == n - 1) : print(pre[i - 1], end = " "); else : print(max(pre[i - 1], suf[i + 1]), end = " ");# Driver code if __name__ == "__main__" : a = [ 2, 5, 6, 1, 3 ]; n = len(a); max_element(a, n); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG {// Function to find maximum element// among (N - 1) elements other than// a[i] for each i from 1 to Nstatic void max_element(int []a, int n){ // To store prefix max element int []pre = new int[n]; pre[0] = a[0]; for (int i = 1; i < n; i++) pre[i] = Math.Max(pre[i - 1], a[i]); // To store suffix max element int []suf = new int[n]; suf[n - 1] = a[n - 1]; for (int i = n - 2; i >= 0; i--) suf[i] = Math.Max(suf[i + 1], a[i]); // Find the maximum element // in the array other than a[i] for (int i = 0; i < n; i++) { if (i == 0) Console.Write(suf[i + 1] + " "); else if (i == n - 1) Console.Write(pre[i - 1] + " "); else Console.Write(Math.Max(pre[i - 1], suf[i + 1]) + " "); }}// Driver codepublic static void Main(String []args) { int []a = { 2, 5, 6, 1, 3 }; int n = a.Length; max_element(a, n);}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript implementation of the approach// Function to find maximum element// among (N - 1) elements other than// a[i] for each i from 1 to Nfunction max_element(a, n){ // To store prefix max element let pre = new Array(n); pre[0] = a[0]; for (let i = 1; i < n; i++) pre[i] = Math.max(pre[i - 1], a[i]); // To store suffix max element let suf = new Array(n); suf[n - 1] = a[n - 1]; for (let i = n - 2; i >= 0; i--) suf[i] = Math.max(suf[i + 1], a[i]); // Find the maximum element // in the array other than a[i] for (let i = 0; i < n; i++) { if (i == 0) document.write(suf[i + 1] + " "); else if (i == n - 1) document.write(pre[i - 1] + " "); else document.write(Math.max(pre[i - 1], suf[i + 1]) + " "); }}// Driver codelet a = [2, 5, 6, 1, 3];let n = a.length;max_element(a, n);// This code is contributed by _saurabh_jaiswal</script> |
Output:
6 6 5 6 6
Time Complexity: O(n)
Auxiliary Space: O(n)
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