# Minimum sum after subtracting multiples of k from the elements of the array

Given an integer and an integer array, the task is to find the minimum possible sum of all the elements of the array after they are reduced by subtracting a multiple of from each element (the result must be positive and every element of the array must be equal after this reduction). If the array cannot be reduced then print

Note that an element may or may not be reduced in the final state of the array.

Examples:

Input: arr[] = {2, 3, 4, 5}, K = 1
Output: 4
Subtract 1 form 2, arr[] = {1, 3, 4, 5}
Subtract 2 from 3, arr[] = {1, 1, 4, 5}
Subtract 3 from 4, arr[] = {1, 1, 1, 5}
Subtract 4 from 5 to make arr[] = {1, 1, 1, 1}, thus giving minimum possible sum as 4.
Input: arr[] = {5, 6, 7}, K = 2
Output: -1

Approach: First, the array needs to be sorted as the problem can be solved using the greedy approach

• Sort the array, if arr[0] < 0 then print -1 as every element needs to be ? 0.
• If K == 0 then no element can be reduced further. So in order to have an answer every element of the array must be equal. So the sum of elements is n * arr[0] else print -1.
• Now for the rest of the elements, run a loop from 1 to n and check whether ((arr[i] – arr[0]) % K) == 0 i.e. arr[i] can be reduced to arr[0].
• If the above condition fails for any element, print -1.
• Else if k == 1 then the answer is n i.e. every element will get reduced to 1.
• Else the answer is n * (a[0] % k).

Below is the implementation of the above approach:

## C++

 // C++ program of the above approach #include using namespace std;   // function to calculate minimum sum after transformation int min_sum(int n, int k, int a[]) {       sort(a, a + n);       if (a[0] < 0)         return -1;             // no element can be reduced further     if (k == 0) {           // if all the elements of the array         // are identical         if (a[0] == a[n - 1])             return (n * a[0]);         else             return -1;     }     else {         int f = 0;         for (int i = 1; i < n; i++) {               int p = a[i] - a[0];               // check if a[i] can be reduced to a[0]             if (p % k == 0)                 continue;             else {                 f = 1;                 break;             }         }           // one of the elements cannot be reduced         // to be equal to the other elements         if (f)             return -1;         else {               // if k = 1 then all elements can be reduced to 1             if (k == 1)                 return n;             else                 return (n * (a[0] % k));         }     } }   // Driver code int main() {     int arr[] = { 2, 3, 4, 5 };     int K = 1;     int N = sizeof(arr) / sizeof(arr[0]);     cout << min_sum(N, K, arr);       return 0; }

## Java

 // Java program of the above approach   import java.io.*; import java.util.*; class GFG {       // function to calculate minimum sum after transformation static int min_sum(int n, int k, int a[]) {       Arrays.sort(a);       if (a[0] < 0)         return -1;             // no element can be reduced further     if (k == 0) {           // if all the elements of the array         // are identical         if (a[0] == a[n - 1])             return (n * a[0]);         else             return -1;     }     else {         int f = 0;         for (int i = 1; i < n; i++) {               int p = a[i] - a[0];               // check if a[i] can be reduced to a[0]             if (p % k == 0)                 continue;             else {                 f = 1;                 break;             }         }           // one of the elements cannot be reduced         // to be equal to the other elements         if (f>0)             return -1;         else {               // if k = 1 then all elements can be reduced to 1             if (k == 1)                 return n;             else                 return (n * (a[0] % k));         }     } }   // Driver code         public static void main (String[] args) {             int arr[] = { 2, 3, 4, 5 };     int K = 1;     int N = arr.length;     System.out.println(min_sum(N, K, arr));     } } // This code is contributed by shs..

## Python3

 # Python 3 program of the above approach   # function to calculate minimum sum # after transformation def min_sum(n, k, a):     a.sort(reverse = False)       if (a[0] < 0):         return -1           # no element can be reduced further     if (k == 0):                   # if all the elements of the         # array are identical         if (a[0] == a[n - 1]):             return (n * a[0])         else:             return -1           else:         f = 0         for i in range(1, n, 1):             p = a[i] - a[0]               # check if a[i] can be             # reduced to a[0]             if (p % k == 0):                 continue             else:                 f = 1                 break                   # one of the elements cannot be reduced         # to be equal to the other elements         if (f):             return -1         else:                           # if k = 1 then all elements             # can be reduced to 1             if (k == 1):                 return n             else:                 return (n * (a[0] % k))       # Driver code if __name__ == '__main__':     arr = [2, 3, 4, 5]     K = 1     N = len(arr)     print(min_sum(N, K, arr))   # This code is contributed by # Surendra_Gangwar

## C#

 // C# program of the above approach using System;   class GFG {       // function to calculate minimum // sum after transformation static int min_sum(int n, int k, int[] a) {       Array.Sort(a);       if (a[0] < 0)         return -1;       // no element can be reduced further     if (k == 0)     {           // if all the elements of the array         // are identical         if (a[0] == a[n - 1])             return (n * a[0]);         else             return -1;     }     else     {         int f = 0;         for (int i = 1; i < n; i++)         {             int p = a[i] - a[0];               // check if a[i] can be             // reduced to a[0]             if (p % k == 0)                 continue;             else             {                 f = 1;                 break;             }         }           // one of the elements cannot be reduced         // to be equal to the other elements         if (f > 0)             return -1;         else         {               // if k = 1 then all elements can             // be reduced to 1             if (k == 1)                 return n;             else                 return (n * (a[0] % k));         }     } }   // Driver code public static void Main () {     int[] arr = new int[] { 2, 3, 4, 5 };     int K = 1;     int N = arr.Length;     Console.WriteLine(min_sum(N, K, arr)); } }   // This code is contributed by mits



## Javascript



Output:

4

Time Complexity: O(nlogn)

Auxiliary Space: O(1)

Further Optimizations:

Instead of sorting the array, we can find the minimum element in O(n) time. We can check whether all elements are the same or not also in O(n) time.

The steps involved in this approach are as follows:

1.  Find the minimum element in the array and store it in variable name val.
2.  Now if val < 0 then print -1 as every element needs to be ? 0 and if the minimum element is greater or equal to 0 then every element will be greater than 0.
3.  If K == 0 then no element can be reduced further. So in order to have an answer every element of the array must be equal. So we check this by iterating a loop and if all are equal then the answer will be n * val else print -1.
4.  Now for the rest of the elements, iterate a loop and check whether ((arr[i] â€“ val) % K) == 0 i.e.  arr[i] can be reduced to val.
5.  If the above condition fails for any element, print -1.
6.  Else if k == 1 then the answer is n i.e. every element will get reduced to 1.
7.  Else the answer is n * (val % k).

Below is the code for the above approach:

## C++

 // C++ program of the above approach   #include   using namespace std;   // function to calculate minimum sum after transformation int min_sum(int n, int k, int a[]) {     // Finding minimum element in the array     int val=INT_MAX;     for(int i=0;i

## Java

 // Java program of above approach import java.io.*; import java.util.*;   class GFG{        static int min_sum(int n,int k,int[] a) {     // Finding minimum element in the array     int val=Integer.MAX_VALUE;     for(int i=0;i

## Python3

 # Python3 program of the above approach   # function to calculate minimum sum after transformation def min_sum(n, k, a):       # Finding minimum element in the array     val = 10000000;     for i in range(n):               val = min(a[i], val);             if (val < 0):         return -1;       # no element can be reduced further     if (k == 0) :           # check if all the elements of the array         # are identical or not         for i in range(n):                       if(val != a[i]):                 return -1;                       return (n * val);           else :         f = 0;         for i in range(n):               p = a[i] - val;               # check if a[i] can be reduced to val             if (p % k == 0):                 continue;             else :                 f = 1;                 break;                                   # one of the elements cannot be reduced         # to be equal to the other elements         if (f > 0):             return -1;         else :               # if k = 1 then all elements can be reduced to 1             if (k == 1):                 return n;             else:                 return (n * (val % k));   # Driver code arr = [ 2, 3, 4, 5 ]; K = 1; N = len(arr); print(min_sum(N, K, arr));     # This code is contributed by phasing17

## C#

 // C# program of the above approach using System; class GFG{       static int min_sum(int n, int k, int[] a) {         // Finding minimum element in the array     int val = Int32.MaxValue;     for(int i = 0; i < n; i++)     {         val = Math.Min(a[i],val);     }       if (val < 0)         return -1;       // no element can be reduced further     if (k == 0) {           // check if all the elements of the array         // are identical or not         for(int i = 0; i < n; i++)         {             if(val !=                a[i])             return -1;         }             return (n * val);     }     else {         int f = 0;         for (int i = 0; i < n; i++) {               int p = a[i] - val;               // check if a[i] can be reduced to val             if (p % k == 0)                 continue;             else {                 f = 1;                 break;             }         }           // one of the elements cannot be reduced         // to be equal to the other elements         if (f!=0)             return -1;         else {               // if k = 1 then all elements can be reduced to 1             if (k == 1)                 return n;             else                 return (n * (val % k));         }     } }   // Driver Code public static void Main () {     int[] arr = { 2, 3, 4, 5 };     int K = 1;     int N = arr.Length;       Console.WriteLine(min_sum(N, K, arr)); } }   // This code is contributed by Aman Kumar

## Javascript

 // JavaScript program of the above approach   // function to calculate minimum sum after transformation function min_sum(n, k, a) {     // Finding minimum element in the array     let val = 10000000;     for(var i = 0; i < n; i++)     {         val = Math.min(a[i], val);     }       if (val < 0)         return -1;       // no element can be reduced further     if (k == 0) {           // check if all the elements of the array         // are identical or not         for(var i = 0; i < n; i++)         {             if(val != a[i])             return -1;         }             return (n * val);     }     else {         var f = 0;         for (var i = 0; i < n; i++) {               var p = a[i] - val;               // check if a[i] can be reduced to val             if (p % k == 0)                 continue;             else {                 f = 1;                 break;             }         }           // one of the elements cannot be reduced         // to be equal to the other elements         if (f > 0)             return -1;         else {               // if k = 1 then all elements can be reduced to 1             if (k == 1)                 return n;             else                 return (n * (val % k));         }     } }   // Driver code let arr = [ 2, 3, 4, 5 ]; let K = 1; let N = arr.length; console.log(min_sum(N, K, arr));   // This code is contributed by phasing17

Output

4

Time Complexity: O(n)

Auxiliary Space: O(1)

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