Count number of substrings with numeric value greater than X

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Given a string ‘S’ (composed of digits) and an integer ‘X”, the task is to count all the sub-strings of ‘S’ that satisfy the following conditions:

The sub-string must not begin with the digit ‘0’.
And the numeric number it represents must be greater than ‘X’.

Note: Two ways of selecting a sub-string are different if they begin or end at different indices.

Examples:

Input: S = "471", X = 47
Output: 2
Only the sub-strings "471" and "71" 
satisfy the given conditions.

Input: S = "2222", X = 97
Output: 3
Valid strings are "222", "222" and "2222".

Approach:

Iterate over each digit of the string ‘S’ and choose the digits which are greater than ‘0’.
Now, take all possible sub-strings starting from the character chosen in the previous step and convert each sub-string to an integer.
Compare the integer from the previous step to ‘X’. If the number is greater than ‘X’, then increment the count variable.
Finally, print the value of the count variable.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that counts
// valid sub-strings
int count(string S, int X)
{
    int count = 0;
    const int N = S.length();
    for (int i = 0; i < N; ++i) {
 
        // Only take those numbers
        // that do not start with '0'.
        if (S[i] != '0') {
            for (int len = 1; (i + len) <= N; ++len) {
 
                // converting the sub-string
                // starting from index 'i'
                // and having length 'len' to int
                // and checking if it is greater
                // than X or not
                if (stoi(S.substr(i, len)) > X)
                    count++;
            }
        }
    }
    return count;
}
 
// Driver code
int main()
{
 
    string S = "2222";
    int X = 97;
    cout << count(S, X);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
// Function that counts
// valid sub-strings
static int count(String S, int X)
{
    int count = 0;
    int N = S.length();
    for (int i = 0; i < N; ++i)
    {
 
        // Only take those numbers
        // that do not start with '0'.
        if (S.charAt(i) != '0')
        {
            for (int len = 1; 
                (i + len) <= N; ++len) 
            {
 
                // converting the sub-string
                // starting from index 'i'
                // and having length 'len' to 
                // int and checking if it is 
                // greater than X or not
                int num = Integer.parseInt(S.substring(i, i + len));
                if (num > X)
                    count++;
            }
        }
    }
    return count;
}
 
// Driver code
public static void main(String []args)
{
    String S = "2222";
    int X = 97;
    System.out.println(count(S, X));
}
}
 
// This code is contributed by ihritik

Python3

# Python3 implementation of
# the approach
 
# Function that counts
# valid sub-strings
def countSubStr(S, X):
 
    cnt = 0
    N = len(S)
     
    for i in range(0, N):
 
        # Only take those numbers
        # that do not start with '0'.
        if (S[i] != '0'):
             
            j = 1
            while((j + i) <= N):
 
                # converting the sub-string
                # starting from index 'i'
                # and having length 'len' to 
                # int and checking if it is 
                # greater than X or not
                num = int(S[i : i + j])
             
                if (num > X):
                    cnt = cnt + 1
                 
                j = j + 1
         
    return cnt;
 
# Driver code
S = "2222";
X = 97;
print(countSubStr(S, X))
 
# This code is contributed by ihritik

C#

// C# implementation of the approach
using System;
 
class GFG
{
// Function that counts
// valid sub-strings
static int count(string S, int X)
{
    int count = 0;
    int N = S.Length;
    for (int i = 0; i < N; ++i) 
    {
 
        // Only take those numbers
        // that do not start with '0'.
        if (S[i] != '0') 
        {
            for (int len = 1; 
                (i + len) <= N; ++len) 
            {
 
                // converting the sub-string
                // starting from index 'i'
                // and having length 'len' to int
                // and checking if it is greater
                // than X or not
                int num = Int32.Parse(S.Substring(i, len));
                if (num > X)
                    count++;
            }
        }
    }
    return count;
}
 
// Driver code
public static void Main(String []args)
{
    string S = "2222";
    int X = 97;
    Console.WriteLine(count(S, X));
}
}
 
// This code is contributed by ihritik

PHP

<?php
// PHP implementation of the approach
 
// Function that counts
// valid sub-strings
function countSubStr($S, $X)
{
    $cnt = 0;
    $N = strlen($S);
     
    for ($i = 0; $i < $N; ++$i) 
    {
 
        // Only take those numbers
        // that do not start w$ith '0'.
        if ($S[$i] != '0') 
        {
            for ($len = 1; 
                ($i + $len) <= $N; ++$len) 
            {
 
                // converting the sub-str$ing
                // starting from index 'i'
                // and having length 'len' to int
                // and checking if it is greater
                // than X or not
                $num = intval(substr($S, $i, $len));
             
                if ($num > $X)
                    $cnt++;
            }
        }
    }
    return $cnt;
}
 
// Driver code
$S = "2222";
$X = 97;
echo countSubStr($S, $X);
 
// This code is contributed by ihritik
?>

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function that counts
// valid sub-strings
function count(S, X)
{
    var count = 0;
    var N = S.length;
    for (var i = 0; i < N; ++i) {
 
        // Only take those numbers
        // that do not start with '0'.
        if (S[i] != '0') {
            for (var len = 1; (i + len) <= N; ++len) {
 
                // converting the sub-string
                // starting from index 'i'
                // and having length 'len' to int
                // and checking if it is greater
                // than X or not
                if (parseInt(S.substring(i, i+len)) > X)
                    count++;
            }
        }
    }
    return count;
}
 
// Driver code
var S = "2222";
var X = 97;
document.write( count(S, X));
 
</script>

Output

Complexity Analysis:

Time Complexity: O(N³)
Auxiliary Space: O(1)

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