How Many Xs Solution

In this page, we will solve the How Many Xs problem using different approaches: iterative and mathematical. We will provide the implementation of the solution in JavaScript, TypeScript, Python, Java, and C++.

Problem Description

You are given an integer L, R, and X. Find the number of occurrences of X in all the numbers in the range (L, R) excluding L and R.

Examples

Example 1:

Input: L = 10, R = 20, X = 1
Output: 10
Explanation: The digit 1 appears 11 times in numbers from 10 to 19.

Example 2:

Input: L = 100, R = 120, X = 2
Output: 2
Explanation: The digit 2 appears 3 times in numbers from 100 to 119.

Constraints

• 0 <= L <= R <= 10^9
• 0 <= X <= 9

---

Solution for How Many Xs Problem

Intuition and Approach

The problem can be solved using different approaches such as iterative counting and mathematical properties.

• Iterative
• Mathematical

Approach 1: Iterative

The iterative approach involves iterating through each number in the range and counting the occurrences of the digit X.

Implementation

Live Editor

function countXs() {
  const L = 10;
  const R = 20;
  const X = 1;

  const countDigitOccurrences = (L, R, X) => {
    let count = 0;
    for (let i = L+1; i < R; i++) {
      let num = i;
      while (num > 0) {
        if (num % 10 === X) count++;
        num = Math.floor(num / 10);
      }
    }
    return count;
  };

  const result = countDigitOccurrences(L, R, X);
  return (
    <div>
      <p>
        <b>Input:</b> L = {L}, R = {R}, X = {X}
      </p>
      <p>
        <b>Output:</b> {result}
      </p>
    </div>
  );
}

Result

Loading...

Codes in Different Languages

JavaScript

Written by @manishh12

 function countDigitOccurrences(L, R, X) {
   let count = 0;
   for (let i = L+1; i < R; i++) {
     let num = i;
     while (num > 0) {
       if (num % 10 === X) count++;
       num = Math.floor(num / 10);
     }
   }
   return count;
 }

TypeScript

Written by @manishh12

 function countDigitOccurrences(L: number, R: number, X: number): number {
   let count = 0;
   for (let i = L+1; i < R; i++) {
     let num = i;
     while (num > 0) {
       if (num % 10 === X) count++;
       num = Math.floor(num / 10);
     }
   }
   return count;
 }

Python

Written by @manishh12

 def count_digit_occurrences(L, R, X):
     count = 0
     for i in range(L+1, R):
         num = i
         while num > 0:
             if num % 10 == X:
                 count += 1
             num //= 10
     return count

Java

Written by @manishh12

 class Solution {
     public int countDigitOccurrences(int L, int R, int X) {
         int count = 0;
         for (int i = L+1; i < R; i++) {
             int num = i;
             while (num > 0) {
                 if (num % 10 == X) {
                     count++;
                 }
                 num /= 10;
             }
         }
         return count;
     }
 }

C++

Written by @manishh12

 #include <iostream>

 using namespace std;

 int countDigitOccurrences(int L, int R, int X) {
     int count = 0;
     for (int i = L+1; i < R; i++) {
         int num = i;
         while (num > 0) {
             if (num % 10 == X) {
                 count++;
             }
             num /= 10;
         }
     }
     return count;
 }

Complexity Analysis

• Time Complexity: $O((R - L)*logR)$
• Space Complexity: $O(1)$

tip

The mathematical approach is more efficient for large ranges. Choose the method that best suits the given constraints.

References

• GeeksforGeeks Problem: GeeksforGeeks Problem
• Solution Link: How Many X's Solution on GeeksforGeeks
• Authors GeeksforGeeks Profile: Manish Kumar Gupta

---