Given an array arr[] of N integers, the task is to check if it is possible to redistribute the array such that for every 1 ? i ? N (1-based indexing) arr[i] = i. Redistributing the array means all the elements of the array can be changed to any other element but the sum of the resultant array must be equal to the original array sum.
Examples:Â
Input: arr[] = {7, 4, 1, 1, 2}Â
Output: YesÂ
7 + 4 + 1 + 1 + 2 = 15Â
1 + 2 + 3 + 4 + 5 = 15Input: arr[] = {1, 1, 1, 1}Â
Output: NoÂ
1 + 1 + 1 + 1 = 4Â
1 + 2 + 3 + 4 = 10Â
Approach: It is given that the sum of the array must not change after the modification. So, calculate the sum of the given array and in order for the array to be of the form 1, 2, 3, …, N, the sum of the array elements must be (N * (N + 1)) / 2. Else, it is impossible.
Below is the implementation of the above approach:Â
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;Â
// Function that returns true if the// array can be redistributed to// the form 1, 2, 3, ..., Nbool canRedistribute(int* a, int n){Â
    // Calculate the sum of the array elements    int sum = 0;    for (int i = 0; i < n; i++)        sum += a[i];Â
    // If can be redistributed    if (sum == (n * (n + 1)) / 2)        return true;Â
    return false;}Â
// Driver codeint main(){Â Â Â Â int a[] = { 7, 4, 1, 1, 2 };Â Â Â Â int n = sizeof(a) / sizeof(int);Â
    if (canRedistribute(a, n))        cout << "Yes";    else        cout << "No";Â
    return 0;} |
Java
// Java implementation of the approachimport java.io.*;Â
class GFG {Â
// Function that returns true if the// array can be redistributed to// the form 1, 2, 3, ..., Nstatic boolean canRedistribute(int []a, int n){Â
    // Calculate the sum of the array elements    int sum = 0;    for (int i = 0; i < n; i++)        sum += a[i];Â
    // If can be redistributed    if (sum == (n * (n + 1)) / 2)        return true;Â
    return false;}Â
// Driver codepublic static void main (String[] args) {Â Â Â Â int a[] = { 7, 4, 1, 1, 2 };Â Â Â Â int n = a.length;Â
    if (canRedistribute(a, n))        System.out.print( "Yes");    else        System.out.print("No");}}Â
// This code is contributed by anuj_67.. |
Python3
# Python implementation of the approachÂ
# Function that returns true if the# array can be redistributed to# the form 1, 2, 3, ..., Ndef canRedistribute(a, n):Â
    # Calculate the sum of the array elements    sum = 0;    for i in range(n):        sum += a[i];Â
    # If can be redistributed    if (sum == (n * (n + 1)) / 2):        return True;Â
    return False;Â
# Driver codeÂ
a = [7, 4, 1, 1, 2 ];n = len(a);Â
if (canRedistribute(a, n)):Â Â Â Â print("Yes");else:Â Â Â Â print("No");Â
# This code is contributed by 29AjayKumar |
C#
// C# implementation of the approachusing System;Â Â Â Â Â class GFG {Â
// Function that returns true if the// array can be redistributed to// the form 1, 2, 3, ..., Nstatic Boolean canRedistribute(int []a, int n){Â
    // Calculate the sum of the array elements    int sum = 0;    for (int i = 0; i < n; i++)        sum += a[i];Â
    // If can be redistributed    if (sum == (n * (n + 1)) / 2)        return true;Â
    return false;}Â
// Driver codepublic static void Main (String[] args) {Â Â Â Â int []a = { 7, 4, 1, 1, 2 };Â Â Â Â int n = a.Length;Â
    if (canRedistribute(a, n))        Console.WriteLine( "Yes");    else        Console.WriteLine("No");}}Â
/* This code is contributed by PrinciRaj1992 */ |
Javascript
<script>Â //Javascript implementation of the approachÂ
// Function that returns true if the// array can be redistributed to// the form 1, 2, 3, ..., Nfunction canRedistribute(a, n){Â
    // Calculate the sum of the array elements    var sum = 0;    for (var i = 0; i < n; i++)        sum += a[i];Â
    // If can be redistributed    if (sum == (n * (n + 1)) / 2)        return true;Â
    return false;}Â
var a = [ 7, 4, 1, 1, 2 ];var n = a.length;Â
    if (canRedistribute(a, n))        document.write("Yes");    else       document.write( "No");Â
Â
// This code is contributed by SoumikMondal</script> |
Output
Yes
Time Complexity: O(n), where n is the length of the given array.
Auxiliary Space: O(1)
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