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++ program for super ugly number #include<bits/stdc++.h> 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< int > 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< int > arr = {34, 23, 85, 93};
getMin(arr);
return 0;
} // This code is contributed by mohit kumar 29. |
// 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 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) |
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. |
<script> // Javascript program for super ugly number // Function to remove each digit // from the given integer function remove(N, digit)
{ // Convert into string
var strN = N.toString();
// Stores final string
var ans = "" ;
var i;
// Traverse the string
for (i=0;i<strN.length;i++){
if ((strN.charCodeAt(i) - 48) == digit)
{
continue ;
}
// Append it to the
// final string
ans += strN[i];
}
// Return integer value
return parseInt(ans);
} // Function to find the minimum sum by // removing occurences of each digit function getMin(arr)
{ var minSum = 1000000000;
var i,j;
// Iterate in range [0, 9]
for (i = 0; i < 10; i++)
{
var curSum = 0;
// Traverse the array
for (j=0;j<arr.length;j++){
curSum += remove(arr[j], i);
}
// Update the minimum sum
minSum = Math.min(minSum, curSum);
}
// Print the minimized sum
document.write(minSum);
} /* Driver program to test above functions */ var arr = [34, 23, 85, 93];
getMin(arr);
</script> |
100
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1).