Median of Two Sorted Arrays

Submission Detail

2094 / 2094 test cases passed.
 Status:

Accepted
Runtime: 12 ms
Memory Usage: 6.4 MB
Submitted: 0 minutes ago

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c
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Your runtime beats 82.90 % of c submissions.
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Accepted Solutions Memory Distribution

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Your memory usage beats 72.30 % of c submissions.
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Submitted Code: 0 minutes ago

Language: c

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double findKthSortedArrays(int *nums1, int nums1Size, int *nums2, int nums2Size, int kth)
{
if (nums1Size == 0)
{
return nums2[kth - 1];
}
if (nums2Size == 0)
{
return nums1[kth - 1];
}
if (kth == 1)
{
return nums2[0] > nums1[0] ? nums1[0] : nums2[0];
}
int kth_left_half = kth / 2;
int kth_right_half = kth / 2;
if (nums1Size < kth_left_half)
{
kth_left_half = nums1Size;
}
if (nums2Size < kth_right_half)
{
kth_right_half = nums2Size;
}
if (nums1[kth_left_half - 1] <= nums2[kth_right_half - 1])
{
nums1 += kth_left_half;
nums1Size -= kth_left_half;
kth -= kth_left_half;
}
else if (nums1[kth_left_half - 1] > nums2[kth_right_half - 1])
{
nums2 += kth_right_half;
nums2Size -= kth_right_half;
kth -= kth_right_half;
}
return findKthSortedArrays(nums1, nums1Size, nums2, nums2Size, kth);
}
double findMedianSortedArrays(int *nums1, int nums1Size, int *nums2, int nums2Size)
{
// find (nums1Size+nums2Size)/2 th num
int left_kth = (nums1Size+nums2Size+1)/2;
int right_kth = (nums1Size+nums2Size+2)/2;
return (findKthSortedArrays(nums1,nums1Size,nums2,nums2Size,left_kth)+
findKthSortedArrays(nums1,nums1Size,nums2,nums2Size,right_kth))/2.0;
}

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