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++
// CPP program to construct a distinct element// array with given size, sum, element upper// bound and all elements positive#include <bits/stdc++.h>using namespace std;void printArray(int N, int SUM, int K){ // smallest possible sum int minSum = (N * (N + 1)) / 2; // largest possible sum int maxSum = (N * K) - (N * (N - 1)) / 2; if (minSum > SUM || maxSum < SUM) { printf("Not Possible"); return; } // Creating array with minimum possible // sum. int arr[N + 1]; for (int i = 1; i <= N; i++) arr[i] = i; int sum = minSum; // running the loop from last because the // array is sorted and running from last // will give largest numbers for (int i = N; i >= 1; i--) { // replacing i with K, Note arr[i] = i int x = sum + (K - i); if (x < SUM) { sum = sum + (K - i); arr[i] = K; // can't be incremented further K--; // to maintain uniqueness } else { // directly replacing with a suitable // element to make sum as given sum arr[i] += (SUM - sum); sum = SUM; break; } } for (int i = 1; i <= N; i++) cout << arr[i] << " ";}// Driver codeint main(){ int N = 3, SUM = 15, K = 8; printArray(N, SUM, K); return 0;} |
Java
// Java program to construct a distinct element// array with given size, sum, element upper// bound and all elements positiveimport java.io.*;class GFG { static void printArray(int N, int SUM, int K) { // smallest possible sum int minSum = (N * (N + 1)) / 2; // largest possible sum int maxSum = (N * K) - (N * (N - 1)) / 2; if (minSum > SUM || maxSum < SUM) { System.out.println("Not Possible"); return; } // Creating array with // minimum possible sum. int arr[] = new int[N + 1]; for (int i = 1; i <= N; i++) arr[i] = i; int sum = minSum; // running the loop from last because the // array is sorted and running from last // will give largest numbers for (int i = N; i >= 1; i--) { // replacing i with K, Note arr[i] = i int x = sum + (K - i); if (x < SUM) { sum = sum + (K - i); // can't be incremented further arr[i] = K; // to maintain uniqueness K--; } else { // directly replacing with a suitable // element to make sum as given sum arr[i] += (SUM - sum); sum = SUM; break; } } for (int i = 1; i <= N; i++) System.out.print(arr[i] + " "); } // Driver code public static void main(String[] args) { int N = 3, SUM = 15, K = 8; printArray(N, SUM, K); }}// This code is contributed by vt_m |
Python3
# Python 3 program to construct a distinct # element array with given size, sum, # element upper bound and all elements # positivedef printArray(N, SUM, K): # smallest possible sum minSum = (N * (N + 1)) / 2 # largest possible sum maxSum = (N * K) - (N * (N - 1)) / 2 if (minSum > SUM or maxSum < SUM): print("Not Possible") return # Creating array with minimum # possible sum. arr = [0 for i in range(N + 1)] for i in range(1, N + 1, 1): arr[i] = i sum = minSum # running the loop from last because # the array is sorted and running # from last will give largest numbers i = N while(i >= 1): # replacing i with K, Note arr[i] = i x = sum + (K - i) if (x < SUM): sum = sum + (K - i) arr[i] = K # can't be incremented further K -= 1 # to maintain uniqueness else: # directly replacing with a suitable # element to make sum as given sum arr[i] += (SUM - sum) sum = SUM break i -= 1 for i in range(1, N + 1, 1): print(int(arr[i]), end = " ")# Driver codeif __name__ == '__main__': N = 3 SUM = 15 K = 8 printArray(N, SUM, K)# This code is contributed by # Surendra_Gangwar |
C#
// C# program to construct a distinct element// array with given size, sum, element upper// bound and all elements positiveusing System;class GFG { static void printArray(int N, int SUM, int K) { // smallest possible sum int minSum = (N * (N + 1)) / 2; // largest possible sum int maxSum = (N * K) - (N * (N - 1)) / 2; if (minSum > SUM || maxSum < SUM) { Console.WriteLine("Not Possible"); return; } // Creating array with // minimum possible sum. int[] arr = new int[N + 1]; for (int i = 1; i <= N; i++) arr[i] = i; int sum = minSum; // running the loop from last because the // array is sorted and running from last // will give largest numbers for (int i = N; i >= 1; i--) { // replacing i with K, Note arr[i] = i int x = sum + (K - i); if (x < SUM) { sum = sum + (K - i); // can't be incremented further arr[i] = K; // to maintain uniqueness K--; } else { // directly replacing with a suitable // element to make sum as given sum arr[i] += (SUM - sum); sum = SUM; break; } } for (int i = 1; i <= N; i++) Console.Write(arr[i] + " "); } // Driver code public static void Main() { int N = 3, SUM = 15, K = 8; printArray(N, SUM, K); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program to construct a distinct element// array with given size, sum, element upper// bound and all elements positivefunction printArray($N, $SUM, $K){ // smallest possible sum $minSum = ($N * ($N + 1)) / 2; // largest possible sum $maxSum = ($N * $K) - ($N * ($N - 1)) / 2; if ($minSum > $SUM || $maxSum < $SUM) { echo"Not Possible"; return; } // Creating array with minimum possible // sum. $arr = array(); for ($i = 1; $i <= $N; $i++) $arr[$i] = $i; $sum = $minSum; // running the loop from last because the // array is sorted and running from last // will give largest numbers for ($i = $N; $i >= 1; $i--) { // replacing i with K, Note arr[i] = i $x = $sum + ($K - $i); if ($x <$SUM) { $sum = $sum + ($K - $i); $arr[$i] =$K; // can't be incremented further $K--; // to maintain uniqueness } else { // directly replacing with a suitable // element to make sum as given sum $arr[$i] += ($SUM - $sum); $sum = $SUM; break; } } for ($i = 1; $i <= $N; $i++) echo $arr[$i] , " ";}// Driver code $N = 3; $SUM = 15;$K = 8; printArray($N, $SUM, $K);// This code is contributed by inder_verma..?> |
Javascript
<script>// JavaScript program to construct a distinct element// array with given size, sum, element upper// bound and all elements positivefunction printArray(N, SUM, K){ // smallest possible sum let minSum = Math.floor((N * (N + 1)) / 2); // largest possible sum let maxSum = (N * K) - Math.floor((N * (N - 1)) / 2); if (minSum > SUM || maxSum < SUM) { document.write("Not Possible"); return; } // Creating array with minimum possible // sum. let arr = new Array(N + 1); for (let i = 1; i <= N; i++) arr[i] = i; let sum = minSum; // running the loop from last because the // array is sorted and running from last // will give largest numbers for (let i = N; i >= 1; i--) { // replacing i with K, Note arr[i] = i let x = sum + (K - i); if (x < SUM) { sum = sum + (K - i); arr[i] = K; // can't be incremented further K--; // to maintain uniqueness } else { // directly replacing with a suitable // element to make sum as given sum arr[i] += (SUM - sum); sum = SUM; break; } } for (let i = 1; i <= N; i++) document.write(arr[i] + " ");}// Driver code let N = 3, SUM = 15, K = 8; printArray(N, SUM, K);// This code is contributed by Surbhi Tyagi.</script> |
Output
1 6 8
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
