Number of strings that satisfy the given condition

Given N strings of equal lengths. The strings contain only digits (1 to 9). The task is to count the number of strings that have an index position such that the digit at this index position is greater than the digits at same index position of all the other strings.

Examples:

Input: arr[] = {“223”, “232”, “112”}
Output: 2
First digit of 1st and 2nd strings are the largest.
Second digit of the string 2nd is the largest.
Third digit of the string 1st is the largest.

Input: arr[] = {“999”, “122”, “111”}
Output: 1

Approach: For each index position, find the maximal digit in that position across all the strings. And store the indices of the string that satisfy the given condition in a set so that the same string isn’t counted twice for different index positions. Finally, return the size of the set.



Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of valid strings
int countStrings(int n, int m, string s[])
{
  
    // Set to store indices of valid strings
    unordered_set<int> ind;
    for (int j = 0; j < m; j++) {
        int mx = 0;
  
        // Find the maximum digit for current position
        for (int i = 0; i < n; i++)
            mx = max(mx, (int)s[i][j] - '0');
  
        // Add indices of all the strings in the set
        // that contain maximal digit
        for (int i = 0; i < n; i++)
            if (s[i][j] - '0' == mx)
                ind.insert(i);
    }
  
    // Return number of strings in the set
    return ind.size();
}
  
// Driver code
int main()
{
    string s[] = { "223", "232", "112" };
    int m = s[0].length();
    int n = sizeof(s) / sizeof(s[0]);
    cout << countStrings(n, m, s);
}

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Java














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// Java implementation of the approach 


import java.util.*; 


  


class GfG  


{ 


  


// Function to return the count of valid strings 


static int countStrings(int n, int m, String s[]) 


{ 


  


    // Set to store indices of valid strings 


    HashSet<Integer> ind = new HashSet<Integer>(); 


    for (int j = 0; j < m; j++) 


    { 


        int mx = 0; 


  


        // Find the maximum digit for current position 


        for (int i = 0; i < n; i++) 


            mx = Math.max(mx, (int)(s[i].charAt(j) - '0')); 


  


        // Add indices of all the strings in the set 


        // that contain maximal digit 


        for (int i = 0; i < n; i++) 


            if (s[i].charAt(j) - '0' == mx) 


                ind.add(i); 


    } 


  


    // Return number of strings in the set 


    return ind.size(); 


} 


  


// Driver code 


public static void main(String[] args)  


{ 


    String s[] = { "223", "232", "112" }; 


    int m = s[0].length(); 


    int n = s.length; 


    System.out.println(countStrings(n, m, s)); 


} 


} 


  


// This code has been contributed by 29AjayKumar 



















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Python3














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# Python3 implementation of the approach 


  


# Function to return the count of 


# valid strings 


def countStrings(n, m, s): 


  


    # Set to store indices of 


    # valid strings 


    ind = dict() 


    for j in range(m): 


        mx = 0


  


        str1 = s[j] 


  


        # Find the maximum digit for  


        # current position 


        for i in range(n): 


            mx = max(mx, int(str1[i])) 


  


        # Add indices of all the strings in  


        # the set that contain maximal digit 


        for i in range(n): 


            if int(str1[i]) == mx: 


                ind[i] = 1


      


    # Return number of strings  


    # in the set 


    return len(ind) 


  


# Driver code 


s = ["223", "232", "112"] 


m = len(s[0]) 


n = len(s) 


print(countStrings(n, m, s)) 


  


# This code is contributed  


# by Mohit Kumar 



















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C#














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// C# implementation of the approach 


using System; 


using System.Collections.Generic; 


  


class GfG  


{ 


  


// Function to return the count of valid strings 


static int countStrings(int n, int m, String[] s) 


{ 


  


    // Set to store indices of valid strings 


    HashSet<int> ind = new HashSet<int>(); 


    for (int j = 0; j < m; j++) 


    { 


        int mx = 0; 


  


        // Find the maximum digit for current position 


        for (int i = 0; i < n; i++) 


            mx = Math.Max(mx, (int)(s[i][j] - '0')); 


  


        // Add indices of all the strings in the set 


        // that contain maximal digit 


        for (int i = 0; i < n; i++) 


            if (s[i][j] - '0' == mx) 


                ind.Add(i); 


    } 


  


    // Return number of strings in the set 


    return ind.Count; 


} 


  


// Driver code 


public static void Main()  


{ 


    String []s = { "223", "232", "112" }; 


    int m = s[0].Length; 


    int n = s.Length; 


    Console.WriteLine(countStrings(n, m, s)); 


} 


} 


  


/* This code contributed by PrinciRaj1992 */



















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Output:


2





Time Complexity: O(N * M) where N is the number of strings and M is the length of the strings.


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