Construct a distinct elements array with given size, sum and element upper bound
Last Updated :
11 Oct, 2022
Given N, size of the original array, SUM total sum of all elements present in the array and K such that no element in array is greater than K, construct the original array where all elements in the array are unique. If there is no solution, print “Not Possible”.
Note: All elements should be positive.
Examples:
Input: N = 3, SUM = 15, K = 8
Output: array[] = {1, 6, 8}
Explanation: The constructed array has size 3, sum 15 and all elements are smaller than or equal to 8.
Input: N = 2, SUM = 9, K = 10
Output: array[]={1, 8}
Input: N = 3, SUM = 23, K = 8
Output: Not Possible
We must choose N elements and all elements must be positive and no element should be greater than K. Since elements are positive and distinct, smallest possible sum is equal to sum of first N natural numbers which is N * (N + 1)/2.
Since all elements should be smaller than or equal to K, largest possible sum is K + (K-1) + (K-2) + ….. + (K-N+1) = (N*K)- (N*(N-1))/2.
So if the given sum is in between smallest and largest possible sum then only the array can be formed, or else we have to print -1.
Following is the complete algorithm if it is feasible to construct the array.
- Create an array of size N and fill it with first N numbers. So total sum of the array will be smallest possible sum.
- Find the largest element of the array, but as the array is sorted so array[N] will be largest.
- If the largest element is less than K, we replace it with K and check the new sum of the array.
- If it is less than given SUM, we move to N-1 position in the array because array[N] can’t be further incremented and to maintain uniqueness we decrease K by 1.
- If it is greater than given SUM, we replace an element such that sum will be given SUM, and will come out of loop.
- If the largest element is equal to K, we move to N-1 position in the array because array[N] can’t be further incremented and to maintain uniqueness we will decrease K by 1.
- Print the array.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
void printArray( int N, int SUM, int K)
{
int minSum = (N * (N + 1)) / 2;
int maxSum = (N * K) - (N * (N - 1)) / 2;
if (minSum > SUM || maxSum < SUM) {
printf ( "Not Possible" );
return ;
}
int arr[N + 1];
for ( int i = 1; i <= N; i++)
arr[i] = i;
int sum = minSum;
for ( int i = N; i >= 1; i--) {
int x = sum + (K - i);
if (x < SUM) {
sum = sum + (K - i);
arr[i] = K;
K--;
}
else {
arr[i] += (SUM - sum);
sum = SUM;
break ;
}
}
for ( int i = 1; i <= N; i++)
cout << arr[i] << " " ;
}
int main()
{
int N = 3, SUM = 15, K = 8;
printArray(N, SUM, K);
return 0;
}
|
Java
import java.io.*;
class GFG {
static void printArray( int N, int SUM, int K)
{
int minSum = (N * (N + 1 )) / 2 ;
int maxSum = (N * K) - (N * (N - 1 )) / 2 ;
if (minSum > SUM || maxSum < SUM) {
System.out.println( "Not Possible" );
return ;
}
int arr[] = new int [N + 1 ];
for ( int i = 1 ; i <= N; i++)
arr[i] = i;
int sum = minSum;
for ( int i = N; i >= 1 ; i--) {
int x = sum + (K - i);
if (x < SUM) {
sum = sum + (K - i);
arr[i] = K;
K--;
}
else {
arr[i] += (SUM - sum);
sum = SUM;
break ;
}
}
for ( int i = 1 ; i <= N; i++)
System.out.print(arr[i] + " " );
}
public static void main(String[] args)
{
int N = 3 , SUM = 15 , K = 8 ;
printArray(N, SUM, K);
}
}
|
Python3
def printArray(N, SUM , K):
minSum = (N * (N + 1 )) / 2
maxSum = (N * K) - (N * (N - 1 )) / 2
if (minSum > SUM or maxSum < SUM ):
print ( "Not Possible" )
return
arr = [ 0 for i in range (N + 1 )]
for i in range ( 1 , N + 1 , 1 ):
arr[i] = i
sum = minSum
i = N
while (i > = 1 ):
x = sum + (K - i)
if (x < SUM ):
sum = sum + (K - i)
arr[i] = K
K - = 1
else :
arr[i] + = ( SUM - sum )
sum = SUM
break
i - = 1
for i in range ( 1 , N + 1 , 1 ):
print ( int (arr[i]), end = " " )
if __name__ = = '__main__' :
N = 3
SUM = 15
K = 8
printArray(N, SUM , K)
|
C#
using System;
class GFG {
static void printArray( int N, int SUM, int K)
{
int minSum = (N * (N + 1)) / 2;
int maxSum = (N * K) - (N * (N - 1)) / 2;
if (minSum > SUM || maxSum < SUM) {
Console.WriteLine( "Not Possible" );
return ;
}
int [] arr = new int [N + 1];
for ( int i = 1; i <= N; i++)
arr[i] = i;
int sum = minSum;
for ( int i = N; i >= 1; i--) {
int x = sum + (K - i);
if (x < SUM) {
sum = sum + (K - i);
arr[i] = K;
K--;
}
else {
arr[i] += (SUM - sum);
sum = SUM;
break ;
}
}
for ( int i = 1; i <= N; i++)
Console.Write(arr[i] + " " );
}
public static void Main()
{
int N = 3, SUM = 15, K = 8;
printArray(N, SUM, K);
}
}
|
PHP
<?php
function printArray( $N , $SUM , $K )
{
$minSum = ( $N * ( $N + 1)) / 2;
$maxSum = ( $N * $K ) - ( $N * ( $N - 1)) / 2;
if ( $minSum > $SUM || $maxSum < $SUM ) {
echo "Not Possible" ;
return ;
}
$arr = array ();
for ( $i = 1; $i <= $N ; $i ++)
$arr [ $i ] = $i ;
$sum = $minSum ;
for ( $i = $N ; $i >= 1; $i --) {
$x = $sum + ( $K - $i );
if ( $x < $SUM ) {
$sum = $sum + ( $K - $i );
$arr [ $i ] = $K ;
$K --;
}
else {
$arr [ $i ] += ( $SUM - $sum );
$sum = $SUM ;
break ;
}
}
for ( $i = 1; $i <= $N ; $i ++)
echo $arr [ $i ] , " " ;
}
$N = 3; $SUM = 15; $K = 8;
printArray( $N , $SUM , $K );
?>
|
Javascript
<script>
function printArray(N, SUM, K)
{
let minSum = Math.floor((N * (N + 1)) / 2);
let maxSum = (N * K) - Math.floor((N * (N - 1)) / 2);
if (minSum > SUM || maxSum < SUM) {
document.write( "Not Possible" );
return ;
}
let arr = new Array(N + 1);
for (let i = 1; i <= N; i++)
arr[i] = i;
let sum = minSum;
for (let i = N; i >= 1; i--) {
let x = sum + (K - i);
if (x < SUM) {
sum = sum + (K - i);
arr[i] = K;
K--;
}
else {
arr[i] += (SUM - sum);
sum = SUM;
break ;
}
}
for (let i = 1; i <= N; i++)
document.write(arr[i] + " " );
}
let N = 3, SUM = 15, K = 8;
printArray(N, SUM, K);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
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