# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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.length(); ` `    ``int` `n = ``sizeof``(s) / ``sizeof``(s); ` `    ``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 ind = ``new` `HashSet(); ` `    ``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.Length; ` `    ``int` `n = s.Length; ` `    ``Console.WriteLine(countStrings(n, m, s)); ` `} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

Output:

```2
```

Time Complexity: O(N * M) where N is the number of strings and M is the length of the strings. My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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