Given an array arr[], consisting of N distinct positive integers and an integer K, the task is to find the lexicographically smallest permutation of the array, such that the sum of elements of any prefix of the output array is not equal to the K. If there exists no such permutation, then print “-1“. Otherwise, print the output array.
Examples:
Input: arr[] = {2, 6, 4, 5, 3, 1}, K = 15
Output: 1 2 3 4 6 5
Explanation:
The lexicographically smallest permutation of the given array is, {1, 2, 3, 4, 6, 5} having no prefix of sum equal to 15.Input: arr[]={3, 1, 4, 6}, K = 12
Output: 1 3 4 6
Explanation:
The lexicographically smallest permutation of the given array is, {1, 3, 4, 6} having no prefix of sum equal to 12.
Approach: The problem can be solved by first sorting the array in ascending order and then swapping the last element of the prefix whose sum is equal to K, with the next element. Follow the steps below to solve this problem:
- If the sum of the array is equal to K, then print “-1” as it will be impossible to find any permutation of the array satisfying the conditions.
- Sort the array in ascending order.
- Initialize a variable, say preSum as 0 to store the sum of a prefix.
- Iterate over the range [0, N-2] using the variable i and perform the following steps:
- Finally, after completing the above steps, print the elements of the array, arr[].
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the output array// satisfying given conditionsvoid findpermutation(int arr[], int N, int K){ int sum = 0; // Traverse the array arr[] for (int i = 0; i < N; i++) { sum = sum + arr[i]; } // If sum is equal to K if (sum == K) { cout << -1 << endl; } else { // Sort array in ascending order sort(arr, arr + N); // Stores the sum of a prefix int preSum = 0; // Traverse the array arr[] for (int i = 0; i < N; i++) { // Update the preSum preSum = preSum + arr[i]; // If preSum is equal to K if (preSum == K) { // Swap swap(arr[i], arr[i + 1]); break; } } // Print the array arr[] for (int i = 0; i < N; i++) { cout << arr[i] << " "; } cout << endl; }}// Driver codeint main(){ int arr[] = { 2, 6, 4, 5, 3, 1 }; int N = sizeof(arr) / sizeof(arr[0]); int K = 15; findpermutation(arr, N, K);} |
Java
// Java program for the above approachimport java.util.*; class GFG{// Function to find the output array// satisfying given conditionsstatic void findpermutation(int arr[], int N, int K){ int sum = 0; // Traverse the array arr[] for(int i = 0; i < N; i++) { sum = sum + arr[i]; } // If sum is equal to K if (sum == K) { System.out.println(-1); } else { // Sort array in ascending order Arrays.sort(arr); // Stores the sum of a prefix int preSum = 0; // Traverse the array arr[] for(int i = 0; i < N; i++) { // Update the preSum preSum = preSum + arr[i]; // If preSum is equal to K if (preSum == K) { // Swap swap(arr, i, i + 1); break; } } // Print the array arr[] for(int i = 0; i < N; i++) { System.out.print(arr[i] + " "); } System.out.println( ); }}static int[] swap(int []arr, int i, int j){ int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; return arr;}// Driver codepublic static void main(String[] args){ int arr[] = { 2, 6, 4, 5, 3, 1 }; int N = arr.length; int K = 15; findpermutation(arr, N, K);}}// This code is contributed by shivanisinghss2110 |
Python3
# Python3 program for the above approach# Function to find the output array# satisfying given conditionsdef findpermutation(arr, N, K): sum = 0 # Traverse the array arr[] for i in range(0, N): sum = sum + arr[i] # If sum is equal to K if (sum == K): print("-1") else: # Sort array in ascending order arr.sort() # Stores the sum of a prefix preSum = 0 # Traverse the array arr[] for i in range(0, N): # Update the preSum preSum = preSum + arr[i] # If preSum is equal to K if (preSum == K): # Swap temp = arr[i] arr[i] = arr[i + 1] arr[i + 1] = temp # Print the array arr[] for i in range(0, N): print(arr[i], end = " ")# Driver codearr = [ 2, 6, 4, 5, 3, 1 ]N = len(arr)K = 15findpermutation(arr, N, K)# This code is contributed by amreshkumar3 |
C#
// C# program for the above approachusing System;class GFG{// Function to find the output array// satisfying given conditionsstatic void findpermutation(int []arr, int N, int K){ int sum = 0; // Traverse the array arr[] for(int i = 0; i < N; i++) { sum = sum + arr[i]; } // If sum is equal to K if (sum == K) { Console.WriteLine(-1); } else { // Sort array in ascending order Array.Sort(arr); // Stores the sum of a prefix int preSum = 0; // Traverse the array arr[] for(int i = 0; i < N; i++) { // Update the preSum preSum = preSum + arr[i]; // If preSum is equal to K if (preSum == K) { // Swap swap(arr, i, i + 1); break; } } // Print the array arr[] for(int i = 0; i < N; i++) { Console.WriteLine(arr[i] + " "); } Console.WriteLine( ); }}static int[] swap(int []arr, int i, int j){ int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; return arr;}// Driver codestatic void Main(){ int []arr= { 2, 6, 4, 5, 3, 1 }; int N = arr.Length; int K = 15; findpermutation(arr, N, K);}}// This code is contributed by SoumikMondal |
Javascript
<script>// JavaScript program for the above approach// Function to find the output array// satisfying given conditionsfunction findpermutation(arr, N, K){ var sum = 0; var i; // Traverse the array arr[] for (i = 0; i < N; i++) { sum = sum + arr[i]; } // If sum is equal to K if (sum == K) { document.write(-1); } else { // Sort array in ascending order arr = arr.sort(function(a, b) { return a - b; }); // Stores the sum of a prefix var preSum = 0; // Traverse the array arr[] for (i = 0; i < N; i++) { // Update the preSum preSum = preSum + arr[i]; // If preSum is equal to K if (preSum == K) { // Swap var temp = arr[i]; arr[i] = arr[i+1]; arr[i+1] = temp; break; } } // Print the array arr[] for (i = 0; i < N; i++) { document.write(arr[i] + " "); } document.write("<br>") }}// Driver code arr = [2, 6, 4, 5, 3, 1]; N = arr.length; K = 15; findpermutation(arr, N, K);</script> |
Output
1 2 3 4 6 5
Time Complexity: O(N)
Auxiliary Space: O(1)
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