Added 3190-find-minimum-operations-to-make-all-elements-divisible-by-…

dsa-solutions/lc-solutions/3100-3199/3190-find-minimum-operations-to-make-all-elements-divisible-by-three.md

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+---
+id: minimum-operations-to-make-all-elements-divisible-by-three
+title: Find Minimum Operations to Make All Elements Divisible by Three
+level: medium
+sidebar_label: Minimum Operations to Make All Elements Divisible by Three
+tags:
+  - Array
+  - Math
+  - Java
+  - Python
+description: "This document provides solutions for the Find Minimum Operations to Make All Elements Divisible by Three problem."
+---
+
+## Problem Statement
+
+You are given an integer array `nums`. In one operation, you can add or subtract 1 from any element of `nums`.
+
+Return the minimum number of operations to make all elements of `nums` divisible by 3.
+
+**Example 1:**
+
+Input: `nums = [1, 2, 3, 4]`
+Output: `3`
+
+Explanation:
+All array elements can be made divisible by 3 using 3 operations:
+- Subtract 1 from 1.
+- Add 1 to 2.
+- Subtract 1 from 4.
+
+**Example 2:**
+
+Input: `nums = [3, 6, 9]`
+Output: `0`
+
+**Constraints:**
+
+- `1 <= nums.length <= 50`
+- `1 <= nums[i] <= 50`
+
+## Solutions
+
+### Intuition
+
+To make all elements divisible by 3, we need to adjust each element by either adding or subtracting 1 until its remainder when divided by 3 is zero. Each adjustment operation (either adding or subtracting 1) counts as one operation.
+
+### Approach
+
+1. Iterate through each element in the array.
+2. For each element, check its remainder when divided by 3.
+3. If the remainder is 1 or 2, it means we need one operation to make it divisible by 3 (subtract 1 if the remainder is 1, or add 1 if the remainder is 2).
+4. Sum up the number of operations required for each element.
+
+### Implementation
+
+The provided code effectively implements the above approach. Here's the breakdown:
+
+1. **Iteration and Remainder Check:**
+   - Iterate through each element in the `nums` array.
+   - Check the remainder of each element when divided by 3.
+   - Increment the operations counter based on the remainder.
+
+### Java 
+
+```java
+class Solution {
+    public int minimumOperations(int[] nums) {
+        int operations = 0;
+        for (int n : nums) {
+            if (n % 3 == 1 || n % 3 == 2) {
+                operations++;
+            }
+        }
+        return operations;
+    }
+}
+```
+### Python
+
+```Python
+class Solution:
+    def minimumOperations(self, nums: List[int]) -> int:
+        operations = 0
+        for n in nums:
+            if n % 3 == 1 or n % 3 == 2:
+                operations += 1
+        return operations
+
+```
+
+### Conclusion 
+This implementation efficiently computes the minimum operations required to make all elements of nums divisible by 3 by iterating through the array and counting adjustments needed based on remainders.