Given an array A[] of size N, the task is to find the minimum sum of numbers required to be added to array elements to convert the array into a permutation of 1 to N. If the array can not be converted to desired permutation, print -1.
Examples:
Input: A[] = {1, 1, 1, 1, 1}
Output: 10
Explanation: Increment A[1] by 1, A[2] by 2, A[3] by 3, A[4] by 4, thus A[] becomes {1, 2, 3, 4, 5}.
Minimum additions required = 1 + 2 + 3 + 4 = 10
Input: A[] = {2, 2, 3}
Output: -1
Approach: The idea is to use sorting. Follow these steps to solve this problem:
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int minimumAdditions( int a[], int n)
{
sort(a, a + n);
int ans = 0;
for ( int i = 0; i < n; i++) {
if ((i + 1) - a[i] < 0) {
return -1;
}
if ((i + 1) - a[i] > 0) {
ans += (i + 1 - a[i]);
}
}
return ans;
}
int main()
{
int A[] = { 1, 1, 1, 1, 1 };
int n = sizeof (A) / sizeof (A[0]);
cout << minimumAdditions(A, n);
return 0;
}
|
Java
import java.util.Arrays;
public class GFG {
static int minimumAdditions( int a[], int n)
{
Arrays.sort(a);
int ans = 0 ;
for ( int i = 0 ; i < n; i++) {
if ((i + 1 ) - a[i] < 0 ) {
return - 1 ;
}
if ((i + 1 ) - a[i] > 0 ) {
ans += (i + 1 - a[i]);
}
}
return ans;
}
public static void main(String[] args)
{
int A[] = { 1 , 1 , 1 , 1 , 1 };
int n = A.length;
System.out.println(minimumAdditions(A, n));
}
}
|
Python3
def minimumAdditions(a, n):
a = sorted (a)
ans = 0
for i in range (n):
if ((i + 1 ) - a[i] < 0 ):
return - 1
if ((i + 1 ) - a[i] > 0 ):
ans + = (i + 1 - a[i])
return ans
if __name__ = = '__main__' :
A = [ 1 , 1 , 1 , 1 , 1 ]
n = len (A)
print (minimumAdditions(A, n))
|
C#
using System;
class GFG{
static int minimumAdditions( int []a, int n)
{
Array.Sort(a);
int ans = 0;
for ( int i = 0; i < n; i++)
{
if ((i + 1) - a[i] < 0)
{
return -1;
}
if ((i + 1) - a[i] > 0)
{
ans += (i + 1 - a[i]);
}
}
return ans;
}
static void Main()
{
int [] A = { 1, 1, 1, 1, 1 };
int n = A.Length;
Console.Write(minimumAdditions(A, n));
}
}
|
Javascript
<script>
function minimumAdditions(a, n)
{
a.sort( function (a, b){ return a-b;});
let ans = 0;
for (let i = 0; i < n; i++) {
if ((i + 1) - a[i] < 0) {
return -1;
}
if ((i + 1) - a[i] > 0) {
ans += (i + 1 - a[i]);
}
}
return ans;
}
let A = [ 1, 1, 1, 1, 1 ];
let n = A.length;
document.write(minimumAdditions(A, n));
</script>
|
Time Complexity: O(N* log(N))
Auxiliary Space: O(1)
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 geeks!