feat/Find Median of Two Sorted Arrays Without Merging

Rudrajiii · Rudrajiii · commit 65cdc7a33929 · 2024-10-28T19:54:11.000+05:30
diff --git a/Data-Structures/Array/findMedianSortedArraysWithoutMerging.js b/Data-Structures/Array/findMedianSortedArraysWithoutMerging.js
@@ -0,0 +1,78 @@
+/**
+ * Function to find the median of two sorted arrays without merging them.
+ * This algorithm efficiently uses binary search on the smaller array to achieve O(log(min(n, m))) complexity.
+ * Sample Problem : https://leetcode.com/problems/median-of-two-sorted-arrays/description/
+ * 
+ * Approach:
+ * 1. Ensure nums1 is the smaller array to reduce the number of binary search operations.
+ * 2. Use binary search to partition both arrays so that elements on the left are less than or equal to elements on the right.
+ * 3. Based on the combined array length, find the median directly by averaging or picking the middle element.
+ *
+ * Time Complexity: O(log(min(n, m))) - where n and m are the lengths of the two arrays.
+ * Space Complexity: O(1) - only a constant amount of space is used.
+ *
+ * Examples:
+ * nums1 = [1, 3], nums2 = [2]
+ * The combined array would be [1, 2, 3] and the median is 2.
+ * 
+ * nums1 = [1, 2], nums2 = [3, 4]
+ * The combined array would be [1, 2, 3, 4] and the median is (2 + 3) / 2 = 2.5.
+ *
+ * @param {number[]} nums1 - First sorted array.
+ * @param {number[]} nums2 - Second sorted array.
+ * @returns {number} - The median of the two sorted arrays.
+ * @throws Will throw an error if the input arrays are not sorted or valid for median calculation.
+ */
+export function findMedianSortedArraysWithoutMerging(nums1, nums2) {
+    // Checks if array are sorted or not
+    const isSorted = (arr) => {
+        for (let i = 1; i < arr.length; i++) {
+            if (arr[i] < arr[i - 1]) return false;
+        }
+        return true;
+    };
+
+    if (!isSorted(nums1) || !isSorted(nums2)) {
+        throw new Error("Input arrays are not sorted or valid for median calculation.");
+    }
+
+    //First ensure nums1 is the smaller array
+    if (nums1.length > nums2.length) {
+        [nums1, nums2] = [nums2, nums1];
+    }
+
+    const x = nums1.length;
+    const y = nums2.length;
+    let low = 0, high = x;
+
+    while (low <= high) {
+        const partitionX = Math.floor((low + high) / 2);
+        const partitionY = Math.floor((x + y + 1) / 2) - partitionX;
+
+        // Edge values in case of partitions at boundaries
+        const maxX = partitionX === 0 ? -Infinity : nums1[partitionX - 1];
+        const minX = partitionX === x ? Infinity : nums1[partitionX];
+
+        const maxY = partitionY === 0 ? -Infinity : nums2[partitionY - 1];
+        const minY = partitionY === y ? Infinity : nums2[partitionY];
+
+        // Check if partition is correct
+        if (maxX <= minY && maxY <= minX) {
+            // Correct partition found, calculate median
+            const leftMax = Math.max(maxX, maxY);
+            const rightMin = Math.min(minX, minY);
+
+            if ((x + y) % 2 === 0) {
+                return (leftMax + rightMin) / 2;
+            } else {
+                return leftMax;
+            }
+        } else if (maxX > minY) {
+            // Move towards left in nums1
+            high = partitionX - 1; 
+        } else {
+            // Move towards right in nums1
+            low = partitionX + 1;  
+        }
+    }
+}
diff --git a/Data-Structures/Array/test/findMedianSortedArraysWithoutMerging.test.js b/Data-Structures/Array/test/findMedianSortedArraysWithoutMerging.test.js
@@ -0,0 +1,38 @@
+import { findMedianSortedArraysWithoutMerging } from '../findMedianSortedArraysWithoutMerging';
+
+describe('findMedianSortedArrays', () => {
+  it('should find the median of two sorted arrays of equal length', () => {
+    expect(findMedianSortedArraysWithoutMerging([1, 3], [2, 4])).toBe(2.5);
+    expect(findMedianSortedArraysWithoutMerging([1, 2], [3, 4])).toBe(2.5);
+  });
+
+  it('should find the median when arrays have different lengths', () => {
+    expect(findMedianSortedArraysWithoutMerging([1, 3], [2])).toBe(2);
+    expect(findMedianSortedArraysWithoutMerging([1, 2, 3], [4, 5])).toBe(3);
+  });
+
+  it('should find the median when one array is empty', () => {
+    expect(findMedianSortedArraysWithoutMerging([], [1])).toBe(1);
+    expect(findMedianSortedArraysWithoutMerging([], [1, 2, 3])).toBe(2);
+  });
+
+  it('should handle single element arrays', () => {
+    expect(findMedianSortedArraysWithoutMerging([1], [2])).toBe(1.5);
+    expect(findMedianSortedArraysWithoutMerging([1], [2, 3, 4])).toBe(2.5);
+  });
+
+  it('should handle arrays with duplicate elements', () => {
+    expect(findMedianSortedArraysWithoutMerging([1, 2, 2], [2, 3, 4])).toBe(2);
+    expect(findMedianSortedArraysWithoutMerging([1, 1, 1], [1, 1, 1])).toBe(1);
+  });
+
+  it('should throw an error if the arrays are not sorted', () => {
+    expect(() => findMedianSortedArraysWithoutMerging([3, 1], [2, 4])).toThrow('Input arrays are not sorted or valid for median calculation.');
+    expect(() => findMedianSortedArraysWithoutMerging([1, 3, 5], [4, 2])).toThrow('Input arrays are not sorted or valid for median calculation.');
+  });
+
+  it('should work for larger arrays', () => {
+    expect(findMedianSortedArraysWithoutMerging([1, 2, 3, 6, 8], [4, 5, 7, 9])).toBe(5);
+    expect(findMedianSortedArraysWithoutMerging([1, 5, 8, 10], [3, 7, 9, 11, 12])).toBe(8);
+  });
+});
