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Median of Two Sorted Arrays Detailed Explanation - Leetcode Java

Anxious Alpaca answered on November 9, 2022 Popularity 9/10 Helpfulness 1/10

answer Median of Two Sorted Arrays Detailed Explanation - Leetcode Java

related Median of Two Sorted Arrays Leetcode Javascript

Median of Two Sorted Arrays Detailed Explanation - Leetcode Java

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public class MedianOfTwoSortedArrays {

    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        // Check if num1 is smaller than num2
        // If not, then we will swap num1 with num2
        if (nums1.length > nums2.length) {
            return findMedianSortedArrays(nums2, nums1);
        }
        // Lengths of two arrays
        int m = nums1.length;
        int n = nums2.length;
        // Pointers for binary search
        int start = 0;
        int end = m;
        // Binary search starts from here
        while (start <= end) {
            // Partitions of both the array
            int partitionNums1 = (start + end) / 2;
            int partitionNums2 = (m + n + 1) / 2 - partitionNums1;
            // Edge cases
            // If there are no elements left on the left side after partition
            int maxLeftNums1 = partitionNums1 == 0 ? Integer.MIN_VALUE : nums1[partitionNums1 - 1];
            // If there are no elements left on the right side after partition
            int minRightNums1 = partitionNums1 == m ? Integer.MAX_VALUE : nums1[partitionNums1];
            // Similarly for nums2
            int maxLeftNums2 = partitionNums2 == 0 ? Integer.MIN_VALUE : nums2[partitionNums2 - 1];
            int minRightNums2 = partitionNums2 == n ? Integer.MAX_VALUE : nums2[partitionNums2];
            // Check if we have found the match
            if (maxLeftNums1 <= minRightNums2 && maxLeftNums2 <= minRightNums1) {
                // Check if the combined array is of even/odd length
                if ((m + n) % 2 == 0) {
                    return (Math.max(maxLeftNums1, maxLeftNums2) + Math.min(minRightNums1, minRightNums2)) / 2.0;
                } else {
                    return Math.max(maxLeftNums1, maxLeftNums2);
                }
            }
            // If we are too far on the right, we need to go to left side
            else if (maxLeftNums1 > minRightNums2) {
                end = partitionNums1 - 1;
            }
            // If we are too far on the left, we need to go to right side
            else {
                start = partitionNums1 + 1;
            }
        }
        // If we reach here, it means the arrays are not sorted
        throw new IllegalArgumentException();
    }
}

Popularity 9/10 Helpfulness 1/10 Language java

Source: redquark.org

Tags: arrays java median

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Contributed on Dec 05 2022

Anxious Alpaca

0 Answers Avg Quality 2/10

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Median of Two Sorted Arrays Leetcode Javascript

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/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var findMedianSortedArrays = function(nums1, nums2) {

  const length1 = nums1.length
  const length2 = nums2.length
  const totalLength = length1 + length2

  let index1 = 0
  let index2 = 0
  
  let medianValue = [0, 0]
  
  for (let i = 0; i <= Math.floor(totalLength / 2); i++) {
    if (nums1[index1] < nums2[index2] || index2 == length2) {
      medianValue.shift()
      medianValue.push(nums1[index1])
      index1++
    } else {
      medianValue.shift()
      medianValue.push(nums2[index2])
      index2++
    }
  }

  if (totalLength % 2 === 0) {
    medianValue = (medianValue[0] + medianValue[1]) / 2
  } else {
    medianValue = medianValue[1]
  }
  return medianValue
};

Popularity 9/10 Helpfulness 3/10 Language javascript

Source: leetcode.com

Tags: arrays ar

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Contributed on Nov 09 2022

Modern Mantis

0 Answers Avg Quality 2/10

Median of Two Sorted Arrays Detailed Explanation - Leetcode Java

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