# Count of numbers from range [L, R] whose sum of digits is Y

Given an integer Y and a range [L, R], the task is to find the count of all numbers from the given range whose sum of digits is equal to Y.

Examples:

Input: L = 0, R = 11, Y = 2
Output: 2
2 -> 2
11 -> 1 + 1 = 2

Input: L = 500, R = 1000, Y = 6
Output: 3

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

Approach: Initialize count = 0 and for every number from the range, calculate sum of it’s digits. If the digit sum is equal to Y then increment the count. Print the count in the end.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the sum ` `// of digits of n ` `int` `sumOfDigits(``int` `n) ` `{ ` `    ``int` `sum = 0; ` ` `  `    ``while` `(n != 0) { ` ` `  `        ``// Add the last digit ` `        ``// to the sum ` `        ``sum = sum + n % 10; ` ` `  `        ``// Remove the last digit ` `        ``n = n / 10; ` `    ``} ` ` `  `    ``// Return the sum of the digits ` `    ``return` `sum; ` `} ` ` `  `// Function to return the count of numbers ` `// from the range [l, r] whose digit sum ` `// is equal to y ` `int` `countTheNumbers(``int` `l, ``int` `r, ``int` `y) ` `{ ` `    ``// Initialise the count to zero ` `    ``int` `count = 0; ` ` `  `    ``// Loop for each number from the range ` `    ``for` `(``int` `i = l; i <= r; i++) { ` ` `  `        ``// If digit sum = y ` `        ``if` `(sumOfDigits(i) == y) ` `            ``count++; ` `    ``} ` ` `  `    ``// Return the required count ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `l = 0, r = 10, y = 1; ` `    ``cout << countTheNumbers(l, r, y); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GfG  ` `{  ` ` `  `// Function to return the sum  ` `// of digits of n  ` `static` `int` `sumOfDigits(``int` `n)  ` `{  ` `    ``int` `sum = ``0``;  ` ` `  `    ``while` `(n != ``0``)  ` `    ``{  ` ` `  `        ``// Add the last digit  ` `        ``// to the sum  ` `        ``sum = sum + n % ``10``;  ` ` `  `        ``// Remove the last digit  ` `        ``n = n / ``10``;  ` `    ``}  ` ` `  `    ``// Return the sum of the digits  ` `    ``return` `sum;  ` `}  ` ` `  `// Function to return the count of numbers  ` `// from the range [l, r] whose digit sum  ` `// is equal to y  ` `static` `int` `countTheNumbers(``int` `l, ``int` `r, ``int` `y)  ` `{  ` `    ``// Initialise the count to zero  ` `    ``int` `count = ``0``;  ` ` `  `    ``// Loop for each number from the range  ` `    ``for` `(``int` `i = l; i <= r; i++)  ` `    ``{  ` ` `  `        ``// If digit sum = y  ` `        ``if` `(sumOfDigits(i) == y)  ` `            ``count++;  ` `    ``}  ` ` `  `    ``// Return the required count  ` `    ``return` `count;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``int` `l = ``0``, r = ``10``, y = ``1``;  ` `    ``System.out.println(countTheNumbers(l, r, y));  ` `} ` `}  ` ` `  `// This code is contributed by Prerna Saini `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the sum ` `# of digits of n ` `def` `sumOfDigits(n): ` ` `  `    ``Sum` `=` `0` ` `  `    ``while` `(n !``=` `0``): ` ` `  `        ``# Add the last digit to the Sum ` `        ``Sum` `=` `Sum` `+` `n ``%` `10` ` `  `        ``# Remove the last digit ` `        ``n ``=` `n ``/``/` `10` `     `  `    ``# Return the Sum of the digits ` `    ``return` `Sum` ` `  `# Function to return the count of numbers ` `# from the range [l, r] whose digit Sum ` `# is equal to y ` `def` `countTheNumbers(l, r, y): ` ` `  `    ``# Initialise the count to zero ` `    ``count ``=` `0` ` `  `    ``# Loop for each number from the range ` `    ``for` `i ``in` `range``(l, r ``+` `1``): ` ` `  `        ``# If digit sum = y ` `        ``if` `(sumOfDigits(i) ``=``=` `y): ` `            ``count ``+``=` `1` ` `  `    ``# Return the required count ` `    ``return` `count ` ` `  `# Driver code ` `l, r, y ``=` `0``, ``10``, ``1` `print``(countTheNumbers(l, r, y)) ` ` `  `# This code is contributed by mohit kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GfG  ` `{  ` ` `  `// Function to return the sum  ` `// of digits of n  ` `static` `int` `sumOfDigits(``int` `n)  ` `{  ` `    ``int` `sum = 0;  ` ` `  `    ``while` `(n != 0)  ` `    ``{  ` ` `  `        ``// Add the last digit  ` `        ``// to the sum  ` `        ``sum = sum + n % 10;  ` ` `  `        ``// Remove the last digit  ` `        ``n = n / 10;  ` `    ``}  ` ` `  `    ``// Return the sum of the digits  ` `    ``return` `sum;  ` `}  ` ` `  `// Function to return the count of numbers  ` `// from the range [l, r] whose digit sum  ` `// is equal to y  ` `static` `int` `countTheNumbers(``int` `l, ``int` `r, ``int` `y)  ` `{  ` `    ``// Initialise the count to zero  ` `    ``int` `count = 0;  ` ` `  `    ``// Loop for each number from the range  ` `    ``for` `(``int` `i = l; i <= r; i++)  ` `    ``{  ` ` `  `        ``// If digit sum = y  ` `        ``if` `(sumOfDigits(i) == y)  ` `            ``count++;  ` `    ``}  ` ` `  `    ``// Return the required count  ` `    ``return` `count;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main(String[] args)  ` `{  ` `    ``int` `l = 0, r = 10, y = 1;  ` `    ``Console.WriteLine(countTheNumbers(l, r, y));  ` `} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## PHP

 ` `

Output:

```2
```

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