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Solving LeetCode - Median of Two Sorted Arrays

Question

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.

Example 1:

Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.
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Example 2:

Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
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Example 3:

Input: nums1 = [0,0], nums2 = [0,0]
Output: 0.00000
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Example 4:

Input: nums1 = [], nums2 = [1]
Output: 1.00000
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Example 5:

Input: nums1 = [2], nums2 = []
Output: 2.00000
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Constraints:

  • nums1.length == m
  • nums2.length == n
  • 0 <= m <= 1000
  • 0 <= n <= 1000
  • 1 <= m + n <= 2000
  • -106 <= nums1[i], nums2[i] <= 106

Let's Go!

Solve by using PREP.

  • P - Two parameters. Both are an array of integers (nums1 & nums2).
  • R - A number which represents the median of the combined arrays.
  • E - Examples provided by question. (See Above)
  • P - See Below
var findMedianSortedArrays = function(nums1, nums2) {
    // Use Reduce method to combines nums1 & nums2 to single array named data

    // Sort array from smallest to largest number

    // Return the below
    // Conditional if even or odd length data
      // If odd grab middle idx of data
      // Else even grab middle 2 numbers and divide by 2
};

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Translate into code...

var findMedianSortedArrays = function(nums1, nums2) {
    // Use Reduce method to combines nums1 & nums2 to single array named data
    let data = nums1.reduce((result, num) => {
        result.push(num)
        return nums2
    }, nums2)

    // Sort array from smallest to largest number
    data.sort((a,b) => a - b)

    // Return the below
    // Conditional if even or odd length data
      // If odd grab middle idx of data
      // Else even grab middle 2 numbers and divide by 2
    if (data.length % 2 === 0) {
      return (data[data.length/2 - 1] + data[data.length/2]) / 2
    } else {
      return data[data.length / 2 - .5]
    }
};

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Conclusion

& Remember... Happy coding, friends! =)

Sources

Top comments (1)

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ljcdev profile image
ljc-dev

Intersting challenge and nice explanation buddy 😄.