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-stringsint 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 codeint main(){ string S = "2222"; int X = 97; cout << count(S, X); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{// Function that counts// valid sub-stringsstatic 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 codepublic 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-stringsdef 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 codeS = "2222";X = 97;print(countSubStr(S, X))# This code is contributed by ihritik |
C#
// C# implementation of the approachusing System;class GFG{// Function that counts// valid sub-stringsstatic 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 codepublic 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-stringsfunction 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-stringsfunction 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 codevar S = "2222";var X = 97;document.write( count(S, X));</script> |
Output
3
Complexity Analysis:
- Time Complexity: O(N3)
- Auxiliary Space: O(1)
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