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Make all strings from a given array equal by replacing minimum number of characters
  • Difficulty Level : Medium
  • Last Updated : 21 Jan, 2021

Given an array of equal-length strings arr[], the task is to make all the strings of the array equal by replacing any character of a string with any other character, minimum number of times.

Examples:

Input: arr[] = { “west”, “east”, “wait” } 
Output:
Explanation: 
Replacing arr[0][1] with ‘a’ modifies arr[] to { “west”, “east”, “wait” }. 
Replacing arr[1][0] with ‘w’ modifies arr[] to { “wast”, “wast”, “wait” }. 
Replacing arr[2][2] with ‘s’ modifies arr[] to { “wast”, “wast”, “wast” }. 
Therefore, the required output is 3.

Input: arr[] = { “abcd”, “bcde”, “cdef” } 
Output: 8

Approach:The problem can be solved using Hashing. Follow the steps below to solve the problem:



  • Initialize a 2D array, say hash[][], where hash[i][j] stores the frequency of the character i present at the jth index of all the strings.
  • Traverse the array arr[] using variable i. For every ith string encountered, count the frequency of each distinct character of the string and store it into the hash[][] array.
  • Initialize a variable, say cntMinOp, to store the minimum count of operations required to make all the strings of the array equal.
  • Traverse the array hash[][] using variable i. For every ith column encountered, calculate the sum of the column, say Sum, the maximum element in the column, say Max, and update cntMinOp += (Sum – Max).
  • Finally, print the value of cntMinOp.

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum count of
// operations required to make all strings
// equal by replacing characters of strings
int minOperation(string arr[], int N)
{
 
    // Stores minimum count of operations
    // required to make all strings equal
    int cntMinOP = 0;
 
    // Stores length of the string
    int M = arr[0].length();
 
    // hash[i][j]: Stores frequency of character
    // i present at j-th index of all strings
    int hash[256][M];
 
    // Initialize hash[][] to 0
    memset(hash, 0, sizeof(hash));
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // Iterate over characters of
        // current string
        for (int j = 0; j < M; j++) {
 
            // Update frequency of
            // arr[i][j]
            hash[arr[i][j]][j]++;
        }
    }
 
    // Traverse hash[][] array
    for (int i = 0; i < M; i++) {
 
        // Stores sum of i-th column
        int Sum = 0;
 
        // Stores the largest element
        // of i-th column
        int Max = 0;
 
        // Iterate over all possible
        // characters
        for (int j = 0; j < 256; j++) {
 
            // Update Sum
            Sum += hash[j][i];
 
            // Update Max
            Max = max(Max, hash[j][i]);
        }
 
        // Update cntMinOP
        cntMinOP += (Sum - Max);
    }
 
    return cntMinOP;
}
 
// Driver Code
int main()
{
 
    string arr[] = { "abcd", "bcde", "cdef" };
 
    int N = sizeof(arr) / sizeof(arr[0]);
    // Function call
    cout << minOperation(arr, N) << "\n";
}

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Java

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// Java program to implement
// the above approach
import java.util.*;
class GFG
{
 
// Function to find the minimum count of
// operations required to make all Strings
// equal by replacing characters of Strings
static int minOperation(String arr[], int N)
{
 
    // Stores minimum count of operations
    // required to make all Strings equal
    int cntMinOP = 0;
 
    // Stores length of the String
    int M = arr[0].length();
 
    // hash[i][j]: Stores frequency of character
    // i present at j-th index of all Strings
    int [][]hash = new int[256][M];
 
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++)
    {
 
        // Iterate over characters of
        // current String
        for (int j = 0; j < M; j++)
        {
 
            // Update frequency of
            // arr[i][j]
            hash[arr[i].charAt(j)][j]++;
        }
    }
 
    // Traverse hash[][] array
    for (int i = 0; i < M; i++)
    {
 
        // Stores sum of i-th column
        int Sum = 0;
 
        // Stores the largest element
        // of i-th column
        int Max = 0;
 
        // Iterate over all possible
        // characters
        for (int j = 0; j < 256; j++)
        {
 
            // Update Sum
            Sum += hash[j][i];
 
            // Update Max
            Max = Math.max(Max, hash[j][i]);
        }
 
        // Update cntMinOP
        cntMinOP += (Sum - Max);
    }
 
    return cntMinOP;
}
 
// Driver Code
public static void main(String[] args)
{
    String arr[] = { "abcd", "bcde", "cdef" };
    int N = arr.length;
   
    // Function call
    System.out.print(minOperation(arr, N)+ "\n");
}
}
 
// This code is contributed by shikhasingrajput

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Python3

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# Python program to implement
# the above approach
 
# Function to find the minimum count of
# operations required to make all Strings
# equal by replacing characters of Strings
def minOperation(arr, N):
   
    # Stores minimum count of operations
    # required to make all Strings equal
    cntMinOP = 0;
 
    # Stores length of the String
    M = len(arr[0]);
 
    # hash[i][j]: Stores frequency of character
    # i present at j-th index of all Strings
    hash = [[0 for i in range(M)] for j in range(256)];
 
    # Traverse the array arr
    for i in range(N):
 
        # Iterate over characters of
        # current String
        for j in range(M):
           
            # Update frequency of
            # arr[i][j]
            hash[ord(arr[i][j])][j] += 1;
 
    # Traverse hash array
    for i in range(M):
 
        # Stores sum of i-th column
        Sum = 0;
 
        # Stores the largest element
        # of i-th column
        Max = 0;
 
        # Iterate over all possible
        # characters
        for j in range(256):
           
            # Update Sum
            Sum += hash[j][i];
 
            # Update Max
            Max = max(Max, hash[j][i]);
 
        # Update cntMinOP
        cntMinOP += (Sum - Max);
    return cntMinOP;
 
# Driver Code
if __name__ == '__main__':
    arr = ["abcd", "bcde", "cdef"];
    N = len(arr);
 
    # Function call
    print(minOperation(arr, N));
 
    # This code is contributed by 29AjayKumar

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C#

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// C# program to implement
// the above approach
using System;
class GFG
{
 
// Function to find the minimum count of
// operations required to make all Strings
// equal by replacing characters of Strings
static int minOperation(String []arr, int N)
{
 
    // Stores minimum count of operations
    // required to make all Strings equal
    int cntMinOP = 0;
 
    // Stores length of the String
    int M = arr[0].Length;
 
    // hash[i,j]: Stores frequency of character
    // i present at j-th index of all Strings
    int [,]hash = new int[256, M];
 
    // Traverse the array []arr
    for (int i = 0; i < N; i++)
    {
 
        // Iterate over characters of
        // current String
        for (int j = 0; j < M; j++)
        {
 
            // Update frequency of
            // arr[i,j]
            hash[arr[i][j], j]++;
        }
    }
 
    // Traverse hash[,] array
    for (int i = 0; i < M; i++)
    {
 
        // Stores sum of i-th column
        int Sum = 0;
 
        // Stores the largest element
        // of i-th column
        int Max = 0;
 
        // Iterate over all possible
        // characters
        for (int j = 0; j < 256; j++)
        {
 
            // Update Sum
            Sum += hash[j, i];
 
            // Update Max
            Max = Math.Max(Max, hash[j, i]);
        }
 
        // Update cntMinOP
        cntMinOP += (Sum - Max);
    }
    return cntMinOP;
}
 
// Driver Code
public static void Main(String[] args)
{
    String []arr = { "abcd", "bcde", "cdef" };
    int N = arr.Length;
   
    // Function call
    Console.Write(minOperation(arr, N)+ "\n");
}
}
 
// This code is contributed by 29AjayKumar

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Output: 

8

 

Time Complexity:O(N * (M + 256)), where M is the length of the string 
Auxiliary Space:O(M + 256)

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