Problem 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)).
Solution
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class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
//handle empty arrays
if(nums1.length == 0 && nums2.length ==0)
return 0;
if(nums1.length == 0)
return nums2.length%2 == 0 ? (nums2[nums2.length/2] + nums2[(nums2.length-1)/2] )/2.0 : nums2[nums2.length/2];
if(nums2.length == 0)
return nums1.length%2 == 0 ? (nums1[nums1.length/2] + nums1[(nums1.length-1)/2] )/2.0 : nums1[nums1.length/2];
//ensure nums1 is of smaller length
//we will apply Binary Search on smaller array for better time complexity
if(nums1.length > nums2.length){
int[] temp = nums1;
nums1 = nums2;
nums2 = temp;
}
//init
int n = nums1.length;
int m = nums2.length;
int total = n+m;
int half = total/2; //size of partition needed
//binary search on nums1
int l=0;
int r=nums1.length;
while(true){
//IMPORTANT: https://www.geeksforgeeks.org/python-operators/
//int i = (l+r)/2 will not work. We need to floor value which is equivalent of // in Python
int i = (int)Math.floor((r+l)/2.0);
//half - (number of elements till i) - 1 because 0 indexed
//=> half - (i+1) - 1 = half - i - 2;
int j = half - i - 2;
//get the elements which need to be compared
int nums1Left = (i>=0?nums1[i]:Integer.MIN_VALUE);
int nums1Right = (i+1<nums1.length?nums1[i+1]:Integer.MAX_VALUE);
int nums2Left = (j>=0?nums2[j]:Integer.MIN_VALUE);
int nums2Right = (j+1<nums2.length?nums2[j+1]:Integer.MAX_VALUE);
//valid partition
if(nums1Left <= nums2Right && nums2Left <= nums1Right){
//odd
if(total%2 == 1){
return Math.min(nums1Right, nums2Right);
}//even
else{
return (double)(Math.max(nums1Left, nums2Left) + Math.min(nums1Right, nums2Right)) / 2;
}
//need to take lesser elements in nums1 partition
}else if(nums1Left > nums2Right){
r = i-1;
//else need more elements in nums1 partition
}else{
l = i+1;
}
}
}
}
ANOTHER WAY TO CODE
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class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
if (nums1.length > nums2.length) {
int[] temp = nums1;
nums1 = nums2;
nums2 = temp;
}
int n = nums1.length;
int m = nums2.length;
int total = n + m;
int half = total / 2;
int l = 0;
int r = n;
while (l <= r) {
int i = (l + r) / 2;
int j = half - i;
int aLeft = (i > 0) ? nums1[i - 1] : Integer.MIN_VALUE;
int aRight = (i < n) ? nums1[i] : Integer.MAX_VALUE;
int bLeft = (j > 0) ? nums2[j - 1] : Integer.MIN_VALUE;
int bRight = (j < m) ? nums2[j] : Integer.MAX_VALUE;
if (aLeft <= bRight && bLeft <= aRight) {
if (total % 2 == 1) {
return Math.min(aRight, bRight);
} else {
return (Math.max(aLeft, bLeft) + Math.min(aRight, bRight)) / 2.0;
}
} else if (aLeft > bRight) {
r = i - 1;
} else {
l = i + 1;
}
}
return 0;
}
}