Number of strings that satisfy the given condition

Difficulty Level : Medium
Last Updated : 02 Jun, 2021

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 1^st and 2^nd strings are the largest.
Second digit of the string 2^nd is the largest.
Third digit of the string 1^st 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++

// 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);
}

Java

// 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

Python3

# 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

C#

// 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 */

Javascript

<script>
 
// JavaScript implementation of the approach
 
// Function to return the count of valid strings
function countStrings(n,m,s)
{
    // Set to store indices of valid strings
    let ind = new Set();
    for (let j = 0; j < m; j++)
    {
        let mx = 0;
   
        // Find the maximum digit for current position
        for (let i = 0; i < n; i++)
            mx = Math.max(mx, (s[i][j].charCodeAt(0) -
            '0'.charCodeAt(0)));
   
        // Add indices of all the strings in the set
        // that contain maximal digit
        for (let i = 0; i < n; i++)
            if (s[i][j].charCodeAt(0) - '0'.charCodeAt(0) == mx)
                ind.add(i);
    }
   
    // Return number of strings in the set
    return ind.size;
}
 
// Driver code
let s=[ "223", "232", "112"];
let m = s[0].length;
let n = s.length;
document.write(countStrings(n, m, s));
 
// This code is contributed by patel2127
 
</script>

Output:

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

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Table of Contents

Number of strings that satisfy the given condition

C++

Java

Python3

C#

Javascript

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