Number of Digit One Solution (Leetcode) Added problem number 233

dsa-solutions/lc-solutions/0200-0299/0233 - Number of Digit One.md

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+---
+id: number-of-digit-one
+title: Number of Digit One
+sidebar_label: 233 Number of Digit One
+tags:
+- Dynamic Programming
+- Java
+- Recursion
+- Math
+description: "This document provides a solution where we count the total number of digit $1$ appearing in all non-negative integers less than or equal to n."
+---
+
+## Problem
+
+Given an integer n, count the total number of $digit$ 1 appearing in all non-negative integers less than or equal to n.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: n = 13
+
+Output: 6
+
+```
+**Example 2:**
+```
+Input: n = 0
+
+Output: 0
+
+```
+### Constraints
+
+- `0 <= n <= 10^9`
+
+---
+## Approach
+There are four approaches discussed that helps to obtain the solution:
+
+1. **Dynamic Programming Table**:
+
+   - Initialize **'count'** to $0$, which will store the total count of $1's$.
+
+   - Use **'factor'** to isolate each digit position, starting from the units place and moving to higher places (tens, hundreds, etc.).
+
+3. **Iterative Analysis**:
+
+  - Loop through each digit position using **'factor'**, which starts from $1$ and increases by a factor of $10$ in each iteration.
+
+   - For each position defined by **'factor'**, determine:
+
+     -  **'lowerNumbers'**: Numbers to the right of the current digit.
+
+     -  **'currentDigit'**: The digit at the current position.
+
+     -  **'higherNumbers'**: Numbers to the left of the current digit.
+
+3. **Count Calculation**:
+
+   - If **'currentDigit'** is $0$, then the count of $1's$ contributed by the current digit position comes solely from higher numbers.
+
+   - If **'currentDigit'** is $1$, it includes all $1's$ contributed by higher numbers, plus the $1's$ in the lower numbers up to **'lowerNumbers + 1'**.
+
+   - If **'currentDigit'** is greater than $1$, it includes all $1's$ contributed by higher numbers and the full set of lower numbers for that digit position.
+
+4. **Result**:
+
+   - Return the accumulated **'count'** after processing all digit positions.
+
+## Solution for Number of Digit One
+
+This problem can be solved using dynamic programming. The problem requires to count the total number of digit $1$ appearing in all non-negative integers less than or equal to n.
+
+#### Code in Java
+
+ ```java
+class Solution {
+    public int countDigitOne(int n) {
+        if (n <= 0) return 0;
+
+        int count = 0;
+        for (long factor = 1; factor <= n; factor *= 10) {
+            long lowerNumbers = n - (n / factor) * factor;
+            long currentDigit = (n / factor) % 10;
+            long higherNumbers = n / (factor * 10);
+
+            if (currentDigit == 0) {
+                count += higherNumbers * factor;
+            } else if (currentDigit == 1) {
+                count += higherNumbers * factor + lowerNumbers + 1;
+            } else {
+                count += (higherNumbers + 1) * factor;
+            }
+        }
+
+        return count;
+    }
+
+    public static void main(String[] args) {
+        Solution sol = new Solution();
+
+        // Test cases
+        System.out.println(sol.countDigitOne(13)); 
+        System.out.println(sol.countDigitOne(0));  
+    }
+}
+
+```
+
+### Complexity Analysis
+
+#### Time Complexity: $O(log_{10} n)$
+
+> **Reason**: The time complexity is $O(log_{10} n)$, because we are processing each digit position from the least significant to the most significant, and the number of digit positions is logarithmic relative to the input size.
+
+#### Space Complexity: $O(1)$
+
+> **Reason**: The space complexity is $O(1)$, because we only use a constant amount of extra space regardless of the input size.
+
+# References
+
+- **LeetCode Problem:** [Number of Digit One](https://leetcode.com/problems/number-of-digit-one/description/)
+- **Solution Link:** [Number of Digit One Solution on LeetCode](https://leetcode.com/problems/number-of-digit-one/solutions/)
+- **Authors LeetCode Profile:** [Vivek Vardhan](https://leetcode.com/u/vivekvardhan43862/)