Lexicographically smallest permutation of Array such that prefix sum till any index is not equal to K
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:
- Increment preSum by arr[i].
- If preSum is equal to K, then swap arr[i] and arr[i+1] and then break.
- Finally, after completing the above steps, print the elements of the array, arr[].
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void findpermutation( int arr[], int N, int K)
{
int sum = 0;
for ( int i = 0; i < N; i++) {
sum = sum + arr[i];
}
if (sum == K) {
cout << -1 << endl;
}
else {
sort(arr, arr + N);
int preSum = 0;
for ( int i = 0; i < N; i++) {
preSum = preSum + arr[i];
if (preSum == K) {
swap(arr[i], arr[i + 1]);
break ;
}
}
for ( int i = 0; i < N; i++) {
cout << arr[i] << " " ;
}
cout << endl;
}
}
int main()
{
int arr[] = { 2, 6, 4, 5, 3, 1 };
int N = sizeof (arr) / sizeof (arr[0]);
int K = 15;
findpermutation(arr, N, K);
}
|
Java
import java.util.*;
class GFG{
static void findpermutation( int arr[], int N, int K)
{
int sum = 0 ;
for ( int i = 0 ; i < N; i++)
{
sum = sum + arr[i];
}
if (sum == K)
{
System.out.println(- 1 );
}
else
{
Arrays.sort(arr);
int preSum = 0 ;
for ( int i = 0 ; i < N; i++)
{
preSum = preSum + arr[i];
if (preSum == K)
{
swap(arr, i, i + 1 );
break ;
}
}
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;
}
public static void main(String[] args)
{
int arr[] = { 2 , 6 , 4 , 5 , 3 , 1 };
int N = arr.length;
int K = 15 ;
findpermutation(arr, N, K);
}
}
|
Python3
def findpermutation(arr, N, K):
sum = 0
for i in range ( 0 , N):
sum = sum + arr[i]
if ( sum = = K):
print ( "-1" )
else :
arr.sort()
preSum = 0
for i in range ( 0 , N):
preSum = preSum + arr[i]
if (preSum = = K):
temp = arr[i]
arr[i] = arr[i + 1 ]
arr[i + 1 ] = temp
for i in range ( 0 , N):
print (arr[i], end = " " )
arr = [ 2 , 6 , 4 , 5 , 3 , 1 ]
N = len (arr)
K = 15
findpermutation(arr, N, K)
|
C#
using System;
class GFG{
static void findpermutation( int []arr, int N, int K)
{
int sum = 0;
for ( int i = 0; i < N; i++)
{
sum = sum + arr[i];
}
if (sum == K)
{
Console.WriteLine(-1);
}
else
{
Array.Sort(arr);
int preSum = 0;
for ( int i = 0; i < N; i++)
{
preSum = preSum + arr[i];
if (preSum == K)
{
swap(arr, i, i + 1);
break ;
}
}
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;
}
static void Main()
{
int []arr= { 2, 6, 4, 5, 3, 1 };
int N = arr.Length;
int K = 15;
findpermutation(arr, N, K);
}
}
|
Javascript
<script>
function findpermutation(arr, N, K)
{
var sum = 0;
var i;
for (i = 0; i < N; i++) {
sum = sum + arr[i];
}
if (sum == K) {
document.write(-1);
}
else {
arr = arr.sort( function (a, b) {
return a - b;
});
var preSum = 0;
for (i = 0; i < N; i++) {
preSum = preSum + arr[i];
if (preSum == K) {
var temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
break ;
}
}
for (i = 0; i < N; i++) {
document.write(arr[i] + " " );
}
document.write( "<br>" )
}
}
arr = [2, 6, 4, 5, 3, 1];
N = arr.length;
K = 15;
findpermutation(arr, N, K);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
24 Mar, 2022
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