Given a string consisting of integers 0 to 9.The task is to count the number of substrings which when converted into integer are divisible by 4. Substring may contain leading zeroes.

Examples:

Input : "124" Output : 4 Substrings divisible by 4 are "12", "4", "24", "124" . Input : "04" Output : 3 Substring divisible by 4 are "0", "4", "04" .

**Method 1: (Brute Force)** The idea is to find all the substrings of the given string and check if substring is divisible by 4 or not.

**Time Complexity: O( ).**

**Efficient solution :** A number is divisible by 4 if its last two digits are divisible by 4 and single digit numbers divisible by 4 are 4, 8 and 0 . So, to calculate number of substrings divisible by 4 we first count number of 0’s, 4’s and 8’s in the string .Then, we make all pairs of two consecutive characters and convert it into integer. After converting it into integer we check that whether it is divisible by 4 or not . If it is divisible by 4 then all such substring ending with this last two charcters are divisible by 4 . Now, **number of such substrings are basically the index of 1st character of pair**. To make it more clear, consider string “14532465” then possible pairs are “14”, “45”, “53”, “32”, “24”, “46”, “65” . In these pairs only “32” and “24” when converted into integer are divisible by 4 . Then, substrings ( length >= 2 ) divisible by 4 must end with either “32” or “24” So, number of substrings ending with “32” are “14532”, “4532”, “532”, “32” i.e 4 and index of ‘3’ is also 4 . Similarly, number of substrings ending with “24” are 5 .

Thus we get an O(n) solution. Below is the implementation of this approach .

## C/C++

`// C++ program to count number of substrings ` `// divisible by 4. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `int` `countDivisbleby4(` `char` `s[]) ` `{ ` ` ` `int` `n = ` `strlen` `(s); ` ` ` ` ` `// In the first loop we will count number of ` ` ` `// 0's, 4's and 8's present in the string ` ` ` `int` `count = 0; ` ` ` `for` `(` `int` `i = 0; i < n; ++i) ` ` ` `if` `(s[i] == ` `'4'` `|| s[i] == ` `'8'` `|| s[i] == ` `'0'` `) ` ` ` `count++ ; ` ` ` ` ` `// In second loop we will convert pairs ` ` ` `// of two consecutive characters into ` ` ` `// integer and store it in variable h . ` ` ` `// Then we check whether h is divisible by 4 ` ` ` `// or not . If h is divisible we increases ` ` ` `// the count with ( i + 1 ) as index of ` ` ` `// first character of pair ` ` ` `for` `(` `int` `i = 0; i < n - 1; ++i) { ` ` ` `int` `h = ( s[i] - ` `'0'` `) * 10 + ( s[i+1] - ` `'0'` `); ` ` ` `if` `(h % 4 == 0) ` ` ` `count = count + i + 1 ; ` ` ` `} ` ` ` ` ` `return` `count; ` `} ` ` ` `// Driver code to test above function ` `int` `main() ` `{ ` ` ` `char` `s[] = ` `"124"` `; ` ` ` `cout << countDivisbleby4(s); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to count number of substrings ` `// divisible by 4 ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to count number of substrings ` ` ` `// divisible by 4 ` ` ` `static` `int` `countDivisbleby4(String s) ` ` ` `{ ` ` ` `int` `n = s.length(); ` ` ` ` ` `// In the first loop we will count number of ` ` ` `// 0's, 4's and 8's present in the string ` ` ` `int` `count = ` `0` `; ` ` ` `for` `(` `int` `i = ` `0` `; i < n; ++i) ` ` ` `if` `(s.charAt(i) == ` `'4'` `|| s.charAt(i) == ` `'8'` `|| s.charAt(i) == ` `'0'` `) ` ` ` `count++ ; ` ` ` ` ` `// In second loop we will convert pairs ` ` ` `// of two consecutive characters into ` ` ` `// integer and store it in variable h . ` ` ` `// Then we check whether h is divisible by 4 ` ` ` `// or not . If h is divisible we increases ` ` ` `// the count with ( i + 1 ) as index of ` ` ` `// first character of pair ` ` ` `for` `(` `int` `i = ` `0` `; i < n - ` `1` `; ++i) ` ` ` `{ ` ` ` `int` `h = ( s.charAt(i) - ` `'0'` `) * ` `10` `+ ( s.charAt(i+` `1` `) - ` `'0'` `); ` ` ` `if` `(h % ` `4` `== ` `0` `) ` ` ` `count = count + i + ` `1` `; ` ` ` `} ` ` ` ` ` `return` `count; ` ` ` `} ` ` ` ` ` `// driver program ` ` ` `public` `static` `void` `main (String[] args) ` ` ` `{ ` ` ` `String s = ` `"124"` `; ` ` ` `System.out.println(countDivisbleby4(s)); ` ` ` `} ` `} ` ` ` `// Contributed by Pramod Kumar ` |

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## Python3

`# Python 3 program to count number of substrings ` `# divisible by 4. ` ` ` `def` `countDivisbleby4(s): ` ` ` `n ` `=` `len` `(s) ` ` ` ` ` `# In the first loop we will count number of ` ` ` `# 0's, 4's and 8's present in the string ` ` ` `count ` `=` `0` `; ` ` ` `for` `i ` `in` `range` `(` `0` `,n,` `1` `): ` ` ` `if` `(s[i] ` `=` `=` `'4'` `or` `s[i] ` `=` `=` `'8'` `or` `s[i] ` `=` `=` `'0'` `): ` ` ` `count ` `+` `=` `1` ` ` ` ` `# In second loop we will convert pairs ` ` ` `# of two consecutive characters into ` ` ` `# integer and store it in variable h . ` ` ` `# Then we check whether h is divisible by 4 ` ` ` `# or not . If h is divisible we increases ` ` ` `# the count with ( i + 1 ) as index of ` ` ` `# first character of pair ` ` ` `for` `i ` `in` `range` `(` `0` `,n ` `-` `1` `,` `1` `): ` ` ` `h ` `=` `(` `ord` `(s[i]) ` `-` `ord` `(` `'0'` `)) ` `*` `10` `+` `(` `ord` `(s[i` `+` `1` `]) ` `-` `ord` `(` `'0'` `)) ` ` ` `if` `(h ` `%` `4` `=` `=` `0` `): ` ` ` `count ` `=` `count ` `+` `i ` `+` `1` ` ` ` ` `return` `count ` ` ` `# Driver code to test above function ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `s ` `=` `[` `'1'` `,` `'2'` `,` `'4'` `] ` ` ` `print` `(countDivisbleby4(s)) ` ` ` `# This code is contributed by ` `# Surendra_Gangwar ` |

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## C#

`// C# program to count number of ` `// substrings divisible by 4 ` `using` `System; ` ` ` `public` `class` `GFG ` `{ ` ` ` ` ` `// Function to count number of ` ` ` `// substrings divisible by 4 ` ` ` `static` `int` `countDivisbleby4(` `string` `s) ` ` ` `{ ` ` ` `int` `n = s.Length; ` ` ` ` ` `// In the first loop we will count ` ` ` `// number of 0's, 4's and 8's present ` ` ` `// in the string ` ` ` `int` `count = 0; ` ` ` `for` `(` `int` `i = 0; i < n; ++i) ` ` ` ` ` `if` `(s[i] == ` `'4'` `|| s[i] == ` `'8'` ` ` `|| s[i] == ` `'0'` `) ` ` ` `count++ ; ` ` ` ` ` `// In second loop we will convert pairs ` ` ` `// of two consecutive characters into ` ` ` `// integer and store it in variable h . ` ` ` `// Then we check whether h is divisible ` ` ` `// by 4 or not . If h is divisible, we ` ` ` `// increases the count with ( i + 1 ) ` ` ` `// as index of first character of pair ` ` ` `for` `(` `int` `i = 0; i < n - 1; ++i) ` ` ` `{ ` ` ` `int` `h = (s[i] - ` `'0'` `) * 10 + ` ` ` `(s[i + 1] - ` `'0'` `); ` ` ` `if` `(h % 4 == 0) ` ` ` `count = count + i + 1 ; ` ` ` `} ` ` ` ` ` `return` `count; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `Main () ` ` ` `{ ` ` ` `string` `s = ` `"124"` `; ` ` ` `Console.WriteLine(countDivisbleby4(s)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Sam007 ` |

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## PHP

`<?php ` `// PHP program to count number ` `// of substrings divisible by 4. ` ` ` `function` `countDivisbleby4( ` `$s` `) ` `{ ` ` ` `$n` `= ` `strlen` `(` `$s` `); ` ` ` ` ` `// In the first loop we ` ` ` `// will count number of ` ` ` `// 0's, 4's and 8's present ` ` ` `// in the string ` ` ` `$count` `= 0; ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ++` `$i` `) ` ` ` `if` `(` `$s` `[` `$i` `] == ` `'4'` `or` `$s` `[` `$i` `] == ` `'8'` ` ` `or` `$s` `[` `$i` `] == ` `'0'` `) ` ` ` `$count` `++ ; ` ` ` ` ` `// In second loop we will convert pairs ` ` ` `// of two consecutive characters into ` ` ` `// integer and store it in variable h . ` ` ` `// Then we check whether h is divisible by 4 ` ` ` `// or not . If h is divisible we increases ` ` ` `// the count with ( i + 1 ) as index of ` ` ` `// first character of pair ` ` ` `for` `( ` `$i` `= 0; ` `$i` `< ` `$n` `- 1; ++` `$i` `) ` ` ` `{ ` ` ` `$h` `= ( ` `$s` `[` `$i` `] - ` `'0'` `) * 10 + ` ` ` `( ` `$s` `[` `$i` `+1] - ` `'0'` `); ` ` ` `if` `(` `$h` `% 4 == 0) ` ` ` `$count` `= ` `$count` `+ ` `$i` `+ 1 ; ` ` ` `} ` ` ` ` ` `return` `$count` `; ` `} ` ` ` ` ` `// Driver Code ` ` ` `$s` `= ` `"124"` `; ` ` ` `echo` `countDivisbleby4(` `$s` `); ` ` ` `// This code is contributed by anuj_67. ` `?> ` |

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**Output:**

4

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