Merge pull request #1486 from shreyash3087/add/leetcode-697

Docs: Added Solutions to Leetcode 697

Author: ajay-dhangar

dsa-problems/leetcode-problems/0600-0699.md

@@ -597,7 +597,7 @@ export const problems = [
         "problemName": "697. Degree of an Array",
         "difficulty": "Easy",
         "leetCodeLink": "https://leetcode.com/problems/degree-of-an-array",
-        "solutionLink": "#"
+        "solutionLink": "/dsa-solutions/lc-solutions/0600-0699/degree-of-an-array"
     },
     {
         "problemName": "698. Partition to K Equal Sum Subsets",

dsa-solutions/lc-solutions/0600-0699/0697-degree-of-an-array.md

@@ -0,0 +1,131 @@
+---
+id: degree-of-an-array
+title: Degree of an Array
+sidebar_label: 0697 - Degree of an Array
+tags:
+  - Hash Table
+  - Array
+  - Sliding Window
+description: "This is a solution to the Degree of an Array problem on LeetCode."
+---
+
+## Problem Description
+
+Given a non-empty array of non-negative integers `nums`, the **degree** of this array is defined as the maximum frequency of any one of its elements.
+
+Your task is to find the smallest possible length of a (contiguous) subarray of `nums`, that has the same degree as `nums`.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: nums = [1,2,2,3,1]
+Output: 2
+Explanation: 
+The input array has a degree of 2 because both elements 1 and 2 appear twice.
+Of the subarrays that have the same degree:
+[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
+The shortest length is 2. So return 2.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,2,3,1,4,2]
+Output: 6
+Explanation: 
+The degree is 3 because the element 2 is repeated 3 times.
+So [2,2,3,1,4,2] is the shortest subarray, therefore returning 6.
+```
+
+### Constraints
+
+- `nums.length` will be between 1 and 50,000.
+- `nums[i]` will be an integer between 0 and 49,999.
+
+## Solution for Degree of an Array
+
+### Approach: Left and Right Index
+#### Intuition and Algorithm
+
+An array that has degree `d`, must have some element `x` occur `d` times. If some subarray has the same degree, then some element `x` (that occurred d times), still occurs `d` times. The shortest such subarray would be from the first occurrence of `x` until the last occurrence.
+
+For each element in the given array, let's know `left`, the index of its first occurrence; and `right`, the index of its last occurrence. For example, with `nums = [1,2,3,2,5]` we have `left[2] = 1` and `right[2] = 3`.
+
+Then, for each element `x` that occurs the maximum number of times, `right[x] - left[x] + 1` will be our candidate answer, and we'll take the minimum of those candidates.
+
+## Code in Different Languages
+
+<Tabs>
+<TabItem value="java" label="Java">
+  <SolutionAuthor name="@Shreyash3087"/>
+
+```java
+class Solution {
+    public int findShortestSubArray(int[] nums) {
+        Map<Integer, Integer> left = new HashMap(),
+            right = new HashMap(), count = new HashMap();
+
+        for (int i = 0; i < nums.length; i++) {
+            int x = nums[i];
+            if (left.get(x) == null) {
+                left.put(x, I);
+            }
+            right.put(x, i);
+            count.put(x, count.getOrDefault(x, 0) + 1);
+        }
+
+        int ans = nums.length;
+        int degree = Collections.max(count.values());
+        for (int x: count.keySet()) {
+            if (count.get(x) == degree) {
+                ans = Math.min(ans, right.get(x) - left.get(x) + 1);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+</TabItem>
+<TabItem value="python" label="Python">
+  <SolutionAuthor name="@Shreyash3087"/>
+
+```python
+class Solution(object):
+    def findShortestSubArray(self, nums):
+        left, right, count = {}, {}, {}
+        for i, x in enumerate(nums):
+            if x not in left:
+                left[x] = i
+            right[x] = i
+            count[x] = count.get(x, 0) + 1
+
+        ans = len(nums)
+        degree = max(count.values())
+        for x in count:
+            if count[x] == degree:
+                ans = min(ans, right[x] - left[x] + 1)
+
+        return ans
+```
+</TabItem>
+</Tabs>
+
+## Complexity Analysis
+
+### Time Complexity: $O(N)$
+
+> **Reason**:  where N is the length of `nums`. Every loop is through $O(N)$ items with $O(1)$ work inside the for-block.
+
+### Space Complexity: $O(N)$
+
+> **Reason**: the space used by `left`, `right`, and `count`.
+
+
+## References
+
+- **LeetCode Problem**: [Degree of an Array](https://leetcode.com/problems/degree-of-an-array/description/)
+
+- **Solution Link**: [Degree of an Array](https://leetcode.com/problems/degree-of-an-array/solutions/)