Given an array arr[] of integers, the task is to find the minimum number greater than the maximum element from the array which cannot be formed using the numbers in the array (adding elements to form some other number). If no such element exists then print -1.
Examples:
Input: arr[] = {2, 6, 9}
Output: -1
There is no such number greater than 9
that cannot be formed using 2, 6 and 9.Input: arr[] = {6, 7, 15}
Output: 16
16 is the smallest number greater than 15 that
cannot be formed using 6, 7 and 15.
Approach: The problem is similar to the minimum coin change problem with slight modification. First sort the array in ascending order and find the maximum element max that will be the number present at the last index of the array. Check for the numbers in the range (max, 2 * max) for the answer. If a number in this range cannot be formed using the elements of the array then that number is the answer, but if all the numbers can be formed in this range then there exists no such number that cannot be formed using the elements of the array.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns the minimum// number greater than maximum of the// array that cannot be formed using the// elements of the arrayint findNumber(int arr[], int n){ // Sort the given array sort(arr, arr + n); // Maximum number in the array int max = arr[n - 1]; // table[i] will store the minimum number of // elements from the array to form i int table[(2 * max) + 1]; table[0] = 0; for (int i = 1; i < (2 * max) + 1; i++) table[i] = INT_MAX; int ans = -1; // Calculate the minimum number of elements // from the array required to form // the numbers from 1 to (2 * max) for (int i = 1; i < (2 * max) + 1; i++) { for (int j = 0; j < n; j++) { if (arr[j] <= i) { int res = table[i - arr[j]]; if (res != INT_MAX && res + 1 < table[i]) table[i] = res + 1; } } // If there exists a number greater than // the maximum element of the array that can be // formed using the numbers of array if (i > arr[n - 1] && table[i] == INT_MAX) { ans = i; break; } } return ans;}// Driver codeint main(){ int arr[] = { 6, 7, 15 }; int n = (sizeof(arr) / sizeof(int)); cout << findNumber(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.util.Arrays;class GFG {// Function that returns the minimum// number greater than maximum of the// array that cannot be formed using the// elements of the arraystatic int findNumber(int arr[], int n){ // Sort the given array Arrays.sort(arr); // Maximum number in the array int max = arr[n - 1]; // table[i] will store the minimum number of // elements from the array to form i int table[] = new int[(2 * max) + 1]; table[0] = 0; for (int i = 1; i < (2 * max) + 1; i++) table[i] = Integer.MAX_VALUE; int ans = -1; // Calculate the minimum number of elements // from the array required to form // the numbers from 1 to (2 * max) for (int i = 1; i < (2 * max) + 1; i++) { for (int j = 0; j < n; j++) { if (arr[j] <= i) { int res = table[i - arr[j]]; if (res != Integer.MAX_VALUE && res + 1 < table[i]) table[i] = res + 1; } } // If there exists a number greater than // the maximum element of the array that can be // formed using the numbers of array if (i > arr[n - 1] && table[i] == Integer.MAX_VALUE) { ans = i; break; } } return ans;}// Driver codepublic static void main(String[] args) { int arr[] = { 6, 7, 15 }; int n = arr.length; System.out.println(findNumber(arr, n));}}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python3 implementation of the approach# Function that returns the minimum# number greater than Maximum of the# array that cannot be formed using the# elements of the arraydef findNumber(arr, n): # Sort the given array arr = sorted(arr) # Maximum number in the array Max = arr[n - 1] # table[i] will store the minimum number of # elements from the array to form i table = [10**9 for i in range((2 * Max) + 1)] table[0] = 0 ans = -1 # Calculate the minimum number of elements # from the array required to form # the numbers from 1 to (2 * Max) for i in range(1, 2 * Max + 1): for j in range(n): if (arr[j] <= i): res = table[i - arr[j]] if (res != 10**9 and res + 1 < table[i]): table[i] = res + 1 # If there exists a number greater than # the Maximum element of the array that can be # formed using the numbers of array if (i > arr[n - 1] and table[i] == 10**9): ans = i break return ans# Driver codearr = [6, 7, 15]n = len(arr)print(findNumber(arr, n)) # This code is contributed by mohit kumar |
C#
// C# implementation of the approachusing System;class GFG {// Function that returns the minimum// number greater than maximum of the// array that cannot be formed using the// elements of the arraystatic int findNumber(int[] arr, int n){ // Sort the given array Array.Sort(arr); // Maximum number in the array int max = arr[n - 1]; // table[i] will store the minimum number of // elements from the array to form i int[] table = new int[(2 * max) + 1]; table[0] = 0; for (int i = 1; i < (2 * max) + 1; i++) table[i] = int.MaxValue; int ans = -1; // Calculate the minimum number of elements // from the array required to form // the numbers from 1 to (2 * max) for (int i = 1; i < (2 * max) + 1; i++) { for (int j = 0; j < n; j++) { if (arr[j] <= i) { int res = table[i - arr[j]]; if (res != int.MaxValue && res + 1 < table[i]) table[i] = res + 1; } } // If there exists a number greater than // the maximum element of the array that can be // formed using the numbers of array if (i > arr[n - 1] && table[i] == int.MaxValue) { ans = i; break; } } return ans;}// Driver codepublic static void Main() { int[] arr = { 6, 7, 15 }; int n = arr.Length; Console.WriteLine(findNumber(arr, n));}}/* This code contributed by Code_Mech */ |
PHP
<?PHP// PHP implementation of the approach// Function that returns the minimum// number greater than maximum of the// array that cannot be formed using the// elements of the arrayfunction findNumber($arr, $n){ // Sort the given array sort($arr); // Maximum number in the array $max = $arr[$n - 1]; // table[i] will store the minimum number of // elements from the array to form i $table = array((2 * $max) + 1); $table[0] = 0; for ($i = 1; $i < (2 * $max) + 1; $i++) $table[$i] = PHP_INT_MAX; $ans = -1; // Calculate the minimum number of elements // from the array required to form // the numbers from 1 to (2 * max) for ($i = 1; $i < (2 * $max) + 1; $i++) { for ($j = 0; $j < $n; $j++) { if ($arr[$j] <= $i) { $res = $table[$i - $arr[$j]]; if ($res != PHP_INT_MAX && $res + 1 < $table[$i]) $table[$i] = $res + 1; } } // If there exists a number greater than // the maximum element of the array that can be // formed using the numbers of array if ($i > $arr[$n - 1] && $table[$i] == PHP_INT_MAX) { $ans = $i; break; } } return $ans;}// Driver code{ $arr = array(6, 7, 15 ); $n = sizeof($arr); echo(findNumber($arr, $n));}/* This code contributed by Code_Mech*/ |
Javascript
<script>// Javascript implementation of the approach// Function that returns the minimum// number greater than maximum of the// array that cannot be formed using the// elements of the arrayfunction findNumber(arr, n) { // Sort the given array arr.sort((a, b) => a - b); // Maximum number in the array var max = arr[n - 1]; // table[i] will store the minimum number of // elements from the array to form i var table = Array((2 * max) + 1).fill(0); table[0] = 0; for(i = 1; i < (2 * max) + 1; i++) table[i] = Number.MAX_VALUE; var ans = -1; // Calculate the minimum number of elements // from the array required to form // the numbers from 1 to (2 * max) for(i = 1; i < (2 * max) + 1; i++) { for(j = 0; j < n; j++) { if (arr[j] <= i) { var res = table[i - arr[j]]; if (res != Number.MAX_VALUE && res + 1 < table[i]) table[i] = res + 1; } } // If there exists a number greater than // the maximum element of the array that can be // formed using the numbers of array if (i > arr[n - 1] && table[i] == Number.MAX_VALUE) { ans = i; break; } } return ans;}// Driver codevar arr = [ 6, 7, 15 ];var n = arr.length;document.write(findNumber(arr, n));// This code is contributed by umadevi9616 </script> |
Output:
16
Time Complexity : O(MAX*N+N*log(N))
Auxiliary Space: O(MAX)
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