Merge K Sorted Array problem from gfg is added

dsa-solutions/gfg-solutions/Easy problems/Merge-K-Sorted-Arrays.md

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+---
+id: merge-k-sorted-arrays
+title: Merge K Sorted Arrays
+sidebar_label: Merge-K-Sorted-Arrays
+tags:
+  - Sorting
+  - Algorithms
+description: "This tutorial covers the solution to the Merge K Sorted Arrays problem from the GeeksforGeeks."
+---
+## Problem Description
+Given `k` sorted arrays arranged in the form of a matrix of size `k * k`. The task is to merge them into one sorted array. Return the merged sorted array 
+
+## Examples
+
+**Example 1:**
+
+```
+Input: k = 3, arr[][] = {{1,2,3},{4,5,6},{7,8,9}}
+Output: 1 2 3 4 5 6 7 8 9
+Explanation: Above test case has 3 sorted arrays of size 3, 3, 3 arr[][] = [[1, 2, 3],[4, 5, 6],[7, 8, 9]]. The merged list will be [1, 2, 3, 4, 5, 6, 7, 8, 9].
+```
+
+**Example 2:**
+
+```
+Input: k = 4, arr[][]={{1,2,3,4},{2,2,3,4},{5,5,6,6},{7,8,9,9}}
+Output: 1 2 2 2 3 3 4 4 5 5 6 6 7 8 9 9 
+Explanation: Above test case has 4 sorted arrays of size 4, 4, 4, 4 arr[][] = [[1, 2, 2, 2], [3, 3, 4, 4], [5, 5, 6, 6], [7, 8, 9, 9 ]]. The merged list will be [1, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 8, 9, 9].
+```
+
+## Constraints
+
+* `1 <= k <= 100`
+
+## Problem Explanation
+Given k sorted arrays arranged in the form of a matrix of size k * k. The task is to merge them into one sorted array. Return the merged sorted array
+
+## Code Implementation
+
+### C++ Solution
+
+```cpp
+#include <iostream>
+#include <vector>
+#include <algorithm>
+
+std::vector<int> mergeSortedArrays(std::vector<std::vector<int>> arrays) {
+  std::vector<int> result;
+  for (const auto& array : arrays) {
+    result.insert(result.end(), array.begin(), array.end());
+  }
+  std::sort(result.begin(), result.end());
+  return result;
+}
+
+int main() {
+  std::vector<std::vector<int>> arrays = { { 1, 3, 5 }, { 2, 4, 6 }, { 0, 7, 8 } };
+  std::vector<int> result = mergeSortedArrays(arrays);
+  for (int num : result) {
+    std::cout << num << " ";
+  }
+  return 0;
+}
+
+
+```
+
+```java
+
+
+import java.util.Arrays;
+
+public class Main {
+  public static void main(String[] args) {
+    int[][] arrays = { { 1, 3, 5 }, { 2, 4, 6 }, { 0, 7, 8 } };
+    int[] result = mergeSortedArrays(arrays);
+    System.out.println(Arrays.toString(result));
+  }
+
+  public static int[] mergeSortedArrays(int[][] arrays) {
+    int[] result = new int[arrays.length * arrays[0].length];
+    int index = 0;
+    for (int[] array : arrays) {
+      for (int num : array) {
+        result[index++] = num;
+      }
+    }
+    Arrays.sort(result);
+    return result;
+  }
+}
+
+
+```
+
+```python
+import itertools
+
+def merge_sorted_arrays(arrays):
+  return sorted(itertools.chain(*arrays))
+
+```
+
+```javascript
+
+function mergeSortedArrays(arrays) {
+  return arrays.flat().sort((a, b) => a - b);
+}
+
+
+```
+
+## Solution Logic:
+
+
+1. Flatten the 2D array:
+    - Iterate through each subarray in the 2D array.
+    - Concatenate each subarray to a single array.
+2. Sort the 1D array:
+    - Use a sorting algorithm (e.g. quicksort, mergesort) to sort the array in ascending order.
+3. Return the sorted 1D array:
+    - Return the sorted array as the final result.
+
+## Time Complexity
+
+* The time complexity is $O(n*log(n))$ where n is the total number of elements in the arrays.
+
+## Space Complexity
+
+* $O(n)$ where n is the total number of elements in the arrays.