Median of Two Sorted Arrays of Different Sizes using JavaScript
Given two arrays, our goal is to find the median of two sorted arrays of different sizes by merging both of them. The median is the middle value of the array after dividing it into two halves.
Example:
Input: Arr1: [1,4,5], Arr2: [2,8,7,3]
Output: Median = 4
Explanation: The merged and sorted array is: ar3[] = [1,2,3,4,5,7,8]
The median is 4
Table of Content
Using Binary Search
In this approach, we will be performing binary search on the shorter array so that we can partition it in such a way that the newly combined left half of both the arrays will have elements less than the combined right half. Adjust the partition point until the condition for median gets satisfied. If the combined length of the new array is odd, return the maximum element of the left partition. If it's even, return the average of the maximum element of the left partition and the minimum element of the right partition.
Example: The below example shows how to find Median of two sorted arrays of different sizes using Binary Search on the Smaller Array.
function find_Median_Fun(no1, no2) {
// Checks whether no1 is the smaller array
if (no1.length > no2.length) {
return find_Median_Fun(no2, no1);
}
const m = no1.length;
const n = no2.length;
let low = 0, high = m;
// Perform binary search
while (low <= high) {
// Calculate partition points
const partitionX = Math.floor((low + high) / 2);
const partitionY = Math.floor((m + n + 1) / 2) - partitionX;
// Find elements to the left and
// right of the partition for both arrays
const maxX = partitionX === 0 ?
Number.NEGATIVE_INFINITY :
no1[partitionX - 1];
const maxY = partitionY === 0 ?
Number.NEGATIVE_INFINITY :
no2[partitionY - 1];
const minX = partitionX === m ?
Number.POSITIVE_INFINITY :
no1[partitionX];
const minY = partitionY === n ?
Number.POSITIVE_INFINITY :
no2[partitionY];
// Check if partitions are correct
if (maxX <= minY && maxY <= minX) {
// If the combined length of the arrays is even
// return the average of the middle elements
if ((m + n) % 2 === 0) {
return (Math.max(maxX, maxY) +
Math.min(minX, minY)) / 2;
} else {
// If the combined length of the arrays is odd,
// return the maximum of the middle elements
return Math.max(maxX, maxY);
}
} else if (maxX > minY) {
// If partitionX is too big,
// move partitionX to the left
high = partitionX - 1;
} else {
// If partitionX is too small,
// move partitionX to the right
low = partitionX + 1;
}
}
}
const no1 = [1, 3];
const no2 = [2];
console.log(find_Median_Fun(no1, no2));
function find_Median_Fun(no1, no2) { // Checks whether no1 is the smaller array if (no1.length > no2.length) { return find_Median_Fun(no2, no1); } const m = no1.length; const n = no2.length; let low = 0, high = m; // Perform binary search while (low <= high) { // Calculate partition points const partitionX = Math.floor((low + high) / 2); const partitionY = Math.floor((m + n + 1) / 2) - partitionX; // Find elements to the left and // right of the partition for both arrays const maxX = partitionX === 0 ? Number.NEGATIVE_INFINITY : no1[partitionX - 1]; const maxY = partitionY === 0 ? Number.NEGATIVE_INFINITY : no2[partitionY - 1]; const minX = partitionX === m ? Number.POSITIVE_INFINITY : no1[partitionX]; const minY = partitionY === n ? Number.POSITIVE_INFINITY : no2[partitionY]; // Check if partitions are correct if (maxX <= minY && maxY <= minX) { // If the combined length of the arrays is even // return the average of the middle elements if ((m + n) % 2 === 0) { return (Math.max(maxX, maxY) + Math.min(minX, minY)) / 2; } else { // If the combined length of the arrays is odd, // return the maximum of the middle elements return Math.max(maxX, maxY); } } else if (maxX > minY) { // If partitionX is too big, // move partitionX to the left high = partitionX - 1; } else { // If partitionX is too small, // move partitionX to the right low = partitionX + 1; } }}const no1 = [1, 3];const no2 = [2];console.log(find_Median_Fun(no1, no2));Output
2
Time Complexity: O(log(min(m, n))), where m and n are the lengths of the two input arrays.
Space Complexity: O(1)
Using Two-Pointers Approach
The function find_Median_fun finds the median of two sorted arrays by iterating through them simultaneously with two pointers, selecting elements in ascending order until reaching the middle point of the combined arrays. It determines the median based on whether the total length is even or odd and returns the result accordingly.
Example: The example below shows how to find Median of two sorted arrays of different sizes using JavaScript Using Two-Pointers.
function find_Median_fun(nums1, nums2) {
const totalLength = nums1.length + nums2.length;
const ifEven = totalLength % 2 === 0;
const halfLength = Math.floor(totalLength / 2);
// 2 pointers
let i = 0, j = 0;
let count = 0;
let median = 0, prevMedian = 0;
while (count <= halfLength) {
prevMedian = median;
if (i < nums1.length &&
(j >= nums2.length || nums1[i] < nums2[j])) {
median = nums1[i];
i++;
} else {
median = nums2[j];
j++;
}
count++;
}
// If length is even
if (ifEven) {
return (prevMedian + median) / 2;
} else {
return median;
}
}
const nums1 = [1, 3];
const nums2 = [2];
console.log("Median:",
find_Median_fun(nums1, nums2));
function find_Median_fun(nums1, nums2) { const totalLength = nums1.length + nums2.length; const ifEven = totalLength % 2 === 0; const halfLength = Math.floor(totalLength / 2); // 2 pointers let i = 0, j = 0; let count = 0; let median = 0, prevMedian = 0; while (count <= halfLength) { prevMedian = median; if (i < nums1.length && (j >= nums2.length || nums1[i] < nums2[j])) { median = nums1[i]; i++; } else { median = nums2[j]; j++; } count++; } // If length is even if (ifEven) { return (prevMedian + median) / 2; } else { return median; }}const nums1 = [1, 3];const nums2 = [2];console.log("Median:", find_Median_fun(nums1, nums2));Output
Median: 2
Time Complexity: O(m + n), where m and n are the lengths of the two input arrays.
Space Complexity: O(1)
