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 the 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

2

Time Complexity: O(N * M) where N is the number of strings and M is the length of the strings.
Auxiliary Space: O(N * M), for using set, where N is the number of strings and M is the length of the strings.

Approach: Iterative Index Position Comparison

This problem can be solved by iterating through each index position of the given strings and checking if the digit at that position is greater than the corresponding digit of all other strings. We can use a nested loop to compare each string with every other string.

Steps:

1. Initialize a counter variable to keep track of the number of strings that satisfy the condition.
2. Iterate through each index position of the strings from 0 to length-1.
3. For each index position, iterate through each string and compare the digit at that index position with the corresponding digit of all other strings.
4. If the digit is greater than all other digits at the same index position, increment the counter.
5. Return the counter.

C++

#include <bits/stdc++.h>
using namespace std;

// Function to count the number of strings
int count_strings(vector<string> arr)
{
    int n = arr.size();
    int m = arr[0].size();
    int count = 0;

    // Traverse the array of strings
    for (int i = 0; i < m; i++) {
        int max_digit = arr[0][i] - '0';

        // Traverse each strings
        for (int j = 1; j < n; j++) {
            if (arr[j][i] - '0' > max_digit) {
                max_digit = arr[j][i] - '0';
            }
        }
        if (arr[0][i] - '0' == max_digit) {
            count++;
        }
    }

    // Print the resultant count
    return count;
}

// Driver Code
int main()
{
    vector<string> arr = { "223", "232", "112" };

    // Call the function and print the result
    int result = count_strings(arr);
    cout << result << endl;

    return 0;
}

Java

import java.util.ArrayList;

public class Main {
    public static int countStrings(ArrayList<String> arr)
    {
        int n = arr.size();
        int m = arr.get(0).length();
        int count = 0;

        // Traverse the array of strings
        for (int i = 0; i < m; i++) {
            int maxDigit = arr.get(0).charAt(i) - '0';

            // Traverse each strings
            for (int j = 1; j < n; j++) {
                if (arr.get(j).charAt(i) - '0' > maxDigit) {
                    maxDigit = arr.get(j).charAt(i) - '0';
                }
            }
            if (arr.get(0).charAt(i) - '0' == maxDigit) {
                count++;
            }
        }

        // Return the resultant count
        return count;
    }

    // Driver Code
    public static void main(String[] args)
    {
        ArrayList<String> arr = new ArrayList<>();
        arr.add("223");
        arr.add("232");
        arr.add("112");

        // Call the function and print the result
        int result = countStrings(arr);
        System.out.println(result);
    }
}

Python3

# Python program for the above approach

# Function to count the number of strings
def count_strings(arr):
    n = len(arr)
    m = len(arr[0])
    count = 0

    # Traverse the array of strings
    for i in range(m):
        max_digit = int(arr[0][i])

        # Traverse each strings
        for j in range(1, n):
            if int(arr[j][i]) > max_digit:
                max_digit = int(arr[j][i])
        if int(arr[0][i]) == max_digit:
            count += 1

    # Print the resultant count
    return count


# Driver Code
arr = ["223", "232", "112"]

# Call the function and print the result
result = count_strings(arr)

print(result)

C#

using System;
using System.Collections.Generic;

class GFG
{
    // Function to count the number of strings
    static int CountStrings(List<string> arr)
    {
        int n = arr.Count;
        int m = arr[0].Length;
        int count = 0;

        // Traverse the array of strings
        for (int i = 0; i < m; i++)
        {
            int maxDigit = arr[0][i] - '0';

            // Traverse each string
            for (int j = 1; j < n; j++)
            {
                if (arr[j][i] - '0' > maxDigit)
                {
                    maxDigit = arr[j][i] - '0';
                }
            }
            if (arr[0][i] - '0' == maxDigit)
            {
                count++;
            }
        }

        // Return the resultant count
        return count;
    }

    // Driver Code
    static void Main()
    {
        List<string> arr = new List<string> { "223", "232", "112" };

        // Call the function and print the result
        int result = CountStrings(arr);
        Console.WriteLine(result);
    }
}

Javascript

// Function to count the number of strings
function countStrings(arr) {
    let n = arr.length;
    let m = arr[0].length;
    let count = 0;

    // Traverse the array of strings
    for (let i = 0; i < m; i++) {
        let maxDigit = parseInt(arr[0][i]);

        // Traverse each strings
        for (let j = 1; j < n; j++) {
            if (parseInt(arr[j][i]) > maxDigit) {
                maxDigit = parseInt(arr[j][i]);
            }
        }
        if (parseInt(arr[0][i]) === maxDigit) {
            count++;
        }
    }

    // Print the resultant count
    return count;
}

// Driver Code
    let arr = ["223", "232", "112"];

    // Call the function and print the result
    let result = countStrings(arr);
    console.log(result);

// This code is contributed by shivamgupta0987654321

Output

2

Time Complexity: The time complexity of this solution is O(N*M²), where N is the number of strings and M is the length of each string. 
Auxiliary Space: The auxiliary space used by this solution is O(1)