Question

4. Median of Two Sorted Arrays

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Description

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

The overall run time complexity should be O(log (m+n)).

 

Example 1:

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

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.

 

Constraints:

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

Solutions

Solution 1: Divide and Conquer

The problem requires the time complexity of the algorithm to be $O(\log (m + n))$, so we cannot directly traverse the two arrays, but need to use the binary search method.

If $m + n$ is odd, then the median is the $\left\lfloor\frac{m + n + 1}{2}\right\rfloor$-th number; if $m + n$ is even, then the median is the average of the $\left\lfloor\frac{m + n + 1}{2}\right\rfloor$-th and the $\left\lfloor\frac{m + n + 2}{2}\right\rfloor$-th numbers. In fact, we can unify it as the average of the $\left\lfloor\frac{m + n + 1}{2}\right\rfloor$-th and the $\left\lfloor\frac{m + n + 2}{2}\right\rfloor$-th numbers.

Therefore, we can design a function $f(i, j, k)$, which represents the $k$-th smallest number in the interval $[i, m)$ of array $nums1$ and the interval $[j, n)$ of array $nums2$. The median is the average of $f(0, 0, \left\lfloor\frac{m + n + 1}{2}\right\rfloor)$ and $f(0, 0, \left\lfloor\frac{m + n + 2}{2}\right\rfloor)$.

The implementation idea of the function $f(i, j, k)$ is as follows:

• If $i \geq m$, it means that the interval $[i, m)$ of array $nums1$ is empty, so directly return $nums2[j + k - 1]$;
• If $j \geq n$, it means that the interval $[j, n)$ of array $nums2$ is empty, so directly return $nums1[i + k - 1]$;
• If $k = 1$, it means to find the first number, so just return the minimum of $nums1[i]$ and $nums2[j]$;
• Otherwise, we find the $\left\lfloor\frac{k}{2}\right\rfloor$-th number in the two arrays, denoted as $x$ and $y$. (Note, if a certain array does not have the $\left\lfloor\frac{k}{2}\right\rfloor$-th number, then we regard the $\left\lfloor\frac{k}{2}\right\rfloor$-th number as $+\infty$.) Compare the size of $x$ and $y$:
  • If $x \leq y$, it means that the $\left\lfloor\frac{k}{2}\right\rfloor$-th number of array $nums1$ cannot be the $k$-th smallest number, so we can exclude the interval $[i, i + \left\lfloor\frac{k}{2}\right\rfloor)$ of array $nums1$, and recursively call $f(i + \left\lfloor\frac{k}{2}\right\rfloor, j, k - \left\lfloor\frac{k}{2}\right\rfloor)$.
  • If $x > y$, it means that the $\left\lfloor\frac{k}{2}\right\rfloor$-th number of array $nums2$ cannot be the $k$-th smallest number, so we can exclude the interval $[j, j + \left\lfloor\frac{k}{2}\right\rfloor)$ of array $nums2$, and recursively call $f(i, j + \left\lfloor\frac{k}{2}\right\rfloor, k - \left\lfloor\frac{k}{2}\right\rfloor)$.

The time complexity is $O(\log(m + n))$, and the space complexity is $O(\log(m + n))$. Here, $m$ and $n$ are the lengths of arrays $nums1$ and $nums2$ respectively.

class Solution:
  def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
    def f(i: int, j: int, k: int) -> int:
      if i >= m:
        return nums2[j + k - 1]
      if j >= n:
        return nums1[i + k - 1]
      if k == 1:
        return min(nums1[i], nums2[j])
      p = k // 2
      x = nums1[i + p - 1] if i + p - 1 < m else inf
      y = nums2[j + p - 1] if j + p - 1 < n else inf
      return f(i + p, j, k - p) if x < y else f(i, j + p, k - p)

    m, n = len(nums1), len(nums2)
    a = f(0, 0, (m + n + 1) // 2)
    b = f(0, 0, (m + n + 2) // 2)
    return (a + b) / 2

class Solution {
  private int m;
  private int n;
  private int[] nums1;
  private int[] nums2;

  public double findMedianSortedArrays(int[] nums1, int[] nums2) {
    m = nums1.length;
    n = nums2.length;
    this.nums1 = nums1;
    this.nums2 = nums2;
    int a = f(0, 0, (m + n + 1) / 2);
    int b = f(0, 0, (m + n + 2) / 2);
    return (a + b) / 2.0;
  }

  private int f(int i, int j, int k) {
    if (i >= m) {
      return nums2[j + k - 1];
    }
    if (j >= n) {
      return nums1[i + k - 1];
    }
    if (k == 1) {
      return Math.min(nums1[i], nums2[j]);
    }
    int p = k / 2;
    int x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
    int y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
    return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
  }
}

class Solution {
public:
  double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
    int m = nums1.size(), n = nums2.size();
    function<int(int, int, int)> f = [&](int i, int j, int k) {
      if (i >= m) {
        return nums2[j + k - 1];
      }
      if (j >= n) {
        return nums1[i + k - 1];
      }
      if (k == 1) {
        return min(nums1[i], nums2[j]);
      }
      int p = k / 2;
      int x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
      int y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
      return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
    };
    int a = f(0, 0, (m + n + 1) / 2);
    int b = f(0, 0, (m + n + 2) / 2);
    return (a + b) / 2.0;
  }
};

func findMedianSortedArrays(nums1 []int, nums2 []int) float64 {
  m, n := len(nums1), len(nums2)
  var f func(i, j, k int) int
  f = func(i, j, k int) int {
    if i >= m {
      return nums2[j+k-1]
    }
    if j >= n {
      return nums1[i+k-1]
    }
    if k == 1 {
      return min(nums1[i], nums2[j])
    }
    p := k / 2
    x, y := 1<<30, 1<<30
    if ni := i + p - 1; ni < m {
      x = nums1[ni]
    }
    if nj := j + p - 1; nj < n {
      y = nums2[nj]
    }
    if x < y {
      return f(i+p, j, k-p)
    }
    return f(i, j+p, k-p)
  }
  a, b := f(0, 0, (m+n+1)/2), f(0, 0, (m+n+2)/2)
  return float64(a+b) / 2.0
}

function findMedianSortedArrays(nums1: number[], nums2: number[]): number {
  const m = nums1.length;
  const n = nums2.length;
  const f = (i: number, j: number, k: number): number => {
    if (i >= m) {
      return nums2[j + k - 1];
    }
    if (j >= n) {
      return nums1[i + k - 1];
    }
    if (k == 1) {
      return Math.min(nums1[i], nums2[j]);
    }
    const p = Math.floor(k / 2);
    const x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
    const y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
    return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
  };
  const a = f(0, 0, Math.floor((m + n + 1) / 2));
  const b = f(0, 0, Math.floor((m + n + 2) / 2));
  return (a + b) / 2;
}

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var findMedianSortedArrays = function (nums1, nums2) {
  const m = nums1.length;
  const n = nums2.length;
  const f = (i, j, k) => {
    if (i >= m) {
      return nums2[j + k - 1];
    }
    if (j >= n) {
      return nums1[i + k - 1];
    }
    if (k == 1) {
      return Math.min(nums1[i], nums2[j]);
    }
    const p = Math.floor(k / 2);
    const x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
    const y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
    return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
  };
  const a = f(0, 0, Math.floor((m + n + 1) / 2));
  const b = f(0, 0, Math.floor((m + n + 2) / 2));
  return (a + b) / 2;
};

public class Solution {
  private int m;
  private int n;
  private int[] nums1;
  private int[] nums2;

  public double FindMedianSortedArrays(int[] nums1, int[] nums2) {
    m = nums1.Length;
    n = nums2.Length;
    this.nums1 = nums1;
    this.nums2 = nums2;
    int a = f(0, 0, (m + n + 1) / 2);
    int b = f(0, 0, (m + n + 2) / 2);
    return (a + b) / 2.0;
  }

  private int f(int i, int j, int k) {
    if (i >= m) {
      return nums2[j + k - 1];
    }
    if (j >= n) {
      return nums1[i + k - 1];
    }
    if (k == 1) {
      return Math.Min(nums1[i], nums2[j]);
    }
    int p = k / 2;
    int x = i + p - 1 < m ? nums1[i + p - 1] : 1 << 30;
    int y = j + p - 1 < n ? nums2[j + p - 1] : 1 << 30;
    return x < y ? f(i + p, j, k - p) : f(i, j + p, k - p);
  }
}

class Solution {
  /**
   * @param int[] $nums1
   * @param int[] $nums2
   * @return float
   */

  function findMedianSortedArrays($nums1, $nums2) {
    $arr = array_merge($nums1, $nums2);
    sort($arr);
    $cnt_arr = count($arr);

    if ($cnt_arr % 2) {
      return $arr[$cnt_arr / 2];
    } else {
      return ($arr[intdiv($cnt_arr, 2) - 1] + $arr[intdiv($cnt_arr, 2)]) / 2;
    }
  }
}

import std/[algorithm, sequtils]

proc medianOfTwoSortedArrays(nums1: seq[int], nums2: seq[int]): float =
  var
  fullList: seq[int] = concat(nums1, nums2)
  value: int = fullList.len div 2

  fullList.sort()

  if fullList.len mod 2 == 0:
  result = (fullList[value - 1] + fullList[value]) / 2
  else:
  result = fullList[value].toFloat()

# Driver Code

# var
#   arrA: seq[int] = @[1, 2]
#   arrB: seq[int] = @[3, 4, 5]
# echo medianOfTwoSortedArrays(arrA, arrB)