# Minimize sum of given array by removing all occurrences of a single digit

Last Updated : 22 Feb, 2023

Given an array arr[] of size N, the task is to minimize the sum by removing all the occurrences of a single digit.

Examples:

Input: arr[] = {34, 23, 85, 93}
Output: 100
Explanation: Removing the occurrences of the digit 3 from each element of the array modifies arr[] to {4, 2, 85, 9}. Therefore, minimized sum of the array = 4 + 2 + 85 + 9 = 100.

Input: arr[] = {434, 863, 342, 121}
Output: 293

Approach: The idea is to remove all occurrences of each possible digit ( [0, 9] ) one by one and calculate the sum of the array after removal of each of them. Finally, find the minimum of these sums. Follow the steps below to solve the problem:

• Initialize a variable, say minSum, to store the minimum sum and curSum to store the sum obtained after removing all occurrences of a digit.
• Iterate over the digits in the range [0, 9] and perform the following:
• Traverse the array arr[] and check for the minimum sum by removing every digit.
• After removing the digits from the string, convert the string back to an integer and add it to curSum.
• Update the value of minSum after each iteration.
• Print the value of minSum as the required answer.

Below is the implementation of the above approach:

## C++

 // C++ program for super ugly number #include using namespace std;   // Function to remove each digit // from the given integer int remove(int N, int digit) {       // Convert into string     string strN = to_string(N);       // Stores final string     string ans = "";       // Traverse the string     for (char i:strN)     {         if ((i - '0') == digit)         {             continue;         }           // Append it to the         // final string         ans += i;       }         // Return integer value     return stoi(ans); }   // Function to find the minimum sum by // removing occurences of each digit void getMin(vector arr) {     int minSum = INT_MAX;       // Iterate in range [0, 9]     for (int i = 0; i < 10; i++)     {         int curSum = 0;           // Traverse the array         for (int num :arr)             curSum += remove(num, i);           // Update the minimum sum         minSum = min(minSum, curSum);       }       // Print the minimized sum     cout << minSum; }   /* Driver program to test above functions */ int main() {   vector arr = {34, 23, 85, 93};   getMin(arr);   return 0; }   // This code is contributed by mohit kumar 29.

## Java

 // Java program for the above approach import java.util.*; class GFG {   // Function to remove each digit // from the given integer static int remove(int N, int digit) {       // Convert into string     String strN = String.valueOf(N);       // Stores final string     String ans = "";           // Traverse the string     for (char i:strN.toCharArray())     {         if ((i - '0') == digit)         {             continue;         }           // Append it to the         // final string         ans += i;       }         // Return integer value     return Integer.parseInt(ans); }   // Function to find the minimum sum by // removing occurences of each digit static void getMin(int[] arr) {     int minSum = Integer.MAX_VALUE;       // Iterate in range [0, 9]     for (int i = 0; i < 10; i++)     {         int curSum = 0;           // Traverse the array         for (int num :arr)             curSum += remove(num, i);           // Update the minimum sum         minSum = Math.min(minSum, curSum);       }       // Print the minimized sum     System.out.print(minSum); }     // Driver Code public static void main(String[] args) {     int[] arr = {34, 23, 85, 93};     getMin(arr); } }   // This code is contributed by code_hunt.

## Python3

 # Python3 program for the above approach   # Function to remove each digit # from the given integer def remove(N, digit):       # Convert into string     strN = str(N)       # Stores final string     ans = ''       # Traverse the string     for i in strN:         if int(i) == digit:             continue           # Append it to the         # final string         ans += i       # Return integer value     return int(ans)   # Function to find the minimum sum by # removing occurences of each digit def getMin(arr):     minSum = float('inf')       # Iterate in range [0, 9]     for i in range(10):         curSum = 0           # Traverse the array         for num in arr:             curSum += remove(num, i)           # Update the minimum sum         minSum = min(minSum, curSum)       # Print the minimized sum     print(minSum)     # Given array arr = [34, 23, 85, 93] getMin(arr)

## C#

 using System; public class GFG {     // Function to remove each digit   // from the given integer   static int remove(int N, int digit)   {       // Convert into string     String strN = N.ToString();       // Stores final string     String ans = "";       // Traverse the string     foreach(char i in strN.ToCharArray())     {       if ((i - '0') == digit)       {         continue;       }         // Append it to the       // final string       ans += i;     }       // Return integer value     return Int32.Parse(ans);   }     // Function to find the minimum sum by   // removing occurences of each digit   static void getMin(int[] arr)   {     int minSum = Int32.MaxValue;       // Iterate in range [0, 9]     for (int i = 0; i < 10; i++)     {       int curSum = 0;         // Traverse the array       foreach(int num in arr)         curSum += remove(num, i);         // Update the minimum sum       minSum = Math.Min(minSum, curSum);     }       // Print the minimized sum     Console.WriteLine(minSum);   }     // Driver Code   static public void Main (){       int[] arr = {34, 23, 85, 93};     getMin(arr);   } }   // This code is contributed by Dharanendra L V.

## Javascript



Output:

100

Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1).

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