Solution of 2180 (LC No.) is added

Author: Ishitamukherjee2004

dsa-solutions/lc-solutions/2100-2199/2180-Count-Integers-With-Even-Digit-Sum.md

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+---
+id: count-integers-even-digit-sum
+title: Count Integers With Even Digit Sum
+sidebar_label: 2180-Count-Integers-With-Even-Digit-Sum
+tags:
+  - Math
+  - Simulation
+
+description: The problem no. is 2180. The Problem is to Count Integers With Even Digit Sum
+---
+
+## Problem Description
+Given a positive integer `num`, return the number of positive integers less than or equal to `num` whose digit sums are even.
+
+The digit sum of a positive integer is the sum of all its digits.
+
+
+### Example
+
+**Example 1:**
+
+
+```
+Input: num = 4
+Output: 2
+Explanation:
+The only integers less than or equal to 4 whose digit sums are even are 2 and 4.  
+```
+**Example 2:**
+```
+Input: num = 30
+Output: 14
+Explanation:
+The 14 integers less than or equal to 30 whose digit sums are even are
+2, 4, 6, 8, 11, 13, 15, 17, 19, 20, 22, 24, 26, and 28.
+```
+### Constraints
+
+- `1 <= num <= 1000`
+
+## Solution Approach
+
+### Intuition:
+
+To efficiently Count Integers With Even Digit Sum
+
+
+## Solution Implementation
+
+### Code In Different Languages:
+
+<Tabs>
+  <TabItem value="JavaScript" label="JavaScript" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```javascript
+    
+class Solution {
+  countEven(num) {
+    let cnt = 0;
+    for(let i = 1; i <= num; i++){
+      let sum = 0;
+      if(i < 10){
+        if(i % 2 == 0) cnt++;
+      } else{
+        let temp = i;
+        while(temp > 0){
+          let rev = temp % 10;
+          sum += rev;
+          temp = Math.floor(temp / 10);
+        }
+        if(sum % 2 == 0) cnt++;
+      }
+    }
+    return cnt;
+  }
+}
+
+
+    ```
+
+  </TabItem>
+  <TabItem value="TypeScript" label="TypeScript">
+  <SolutionAuthor name="@Ishitamukherjee2004"/> 
+   ```typescript
+    
+class Solution {
+  countEven(num: number): number {
+    let cnt = 0;
+    for(let i = 1; i <= num; i++){
+      let sum = 0;
+      if(i < 10){
+        if(i % 2 == 0) cnt++;
+      } else{
+        let temp = i;
+        while(temp > 0){
+          let rev = temp % 10;
+          sum += rev;
+          temp = Math.floor(temp / 10);
+        }
+        if(sum % 2 == 0) cnt++;
+      }
+    }
+    return cnt;
+  }
+}
+
+
+    ```
+
+  </TabItem>
+  <TabItem value="Python" label="Python"> 
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```python
+    
+class Solution:
+    def countEven(self, num):
+        cnt = 0
+        for i in range(1, num + 1):
+            sum = 0
+            if i < 10:
+                if i % 2 == 0:
+                    cnt += 1
+            else:
+                temp = i
+                while temp > 0:
+                    rev = temp % 10
+                    sum += rev
+                    temp = temp // 10
+                if sum % 2 == 0:
+                    cnt += 1
+        return cnt
+
+
+    ```
+
+  </TabItem>
+  <TabItem value="Java" label="Java">
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```java
+   public class Solution {
+    public int countEven(int num) {
+        int cnt = 0;
+        for(int i = 1; i <= num; i++){
+            int sum = 0;
+            if(i < 10){
+                if(i % 2 == 0) cnt++;
+            } else{
+                int temp = i;
+                while(temp > 0){
+                    int rev = temp % 10;
+                    sum += rev;
+                    temp = temp / 10;
+                }
+                if(sum % 2 == 0) cnt++;
+            }
+        }
+        return cnt;
+    }
+}
+
+    ```
+
+  </TabItem>
+  <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```cpp
+class Solution {
+public:
+    int countEven(int num) {
+        int cnt = 0;
+        for(int i=1; i<=num; i++){
+            int sum = 0;
+            if(i<10){
+                if(i%2==0) cnt++;
+            }
+            else{
+                int temp = i;
+                while(temp>0){
+                    int rev = temp%10;
+                    sum+=rev;
+                    temp/=10;
+                }
+                if(sum%2==0) cnt++;
+
+            }
+        }
+        return cnt;
+    }
+};
+
+```
+</TabItem> 
+</Tabs>
+
+#### Complexity Analysis
+
+- Time Complexity: $$O(n)$$
+- Space Complexity: $$O(1)$$
+- The time complexity is $$O(n)$$,where n is the input number num. This is because the algorithm uses a single loop that iterates from 1 to num.
+- The space complexity is $$O(1)$$ because we are not using any extra space.