Add 4-Median-of-Two-Sorted-Arrays.c

Create the C solution similar to the YouTube video solution "https://www.youtube.com/watch?v=q6IEA26hvXc"

c/4-Median-of-Two-Sorted-Arrays.c

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+/*
+ Time: log(min(n, m))
+ Space: O(1)
+*/
+
+#define min(x, y) ((x < y) ? (x) : (y))
+#define max(x, y) ((x > y) ? (x) : (y))
+
+double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size){
+ int* A = nums1;
+ int* B = nums2;
+ int ASize = nums1Size;
+ int BSize = nums2Size;
+
+ int total = nums1Size + nums2Size;
+ int half = total / 2;
+
+ if(nums2Size < nums1Size)
+ {
+ A = nums2;
+ B = nums1;
+ ASize = nums2Size;
+ BSize = nums1Size;
+ }
+
+ int l = 0;
+ int r = ASize - 1;
+
+ while(true)
+ {
+ int i = l + ((r - l - 2 + 1) / 2); // A Mid, round down instead of rounding towards 0
+ int j = half - i - 2; // B Mid
+
+
+ int Aleft = i >= 0 ? A[i] : INT_MIN;
+ int Aright = (i + 1) < ASize ? A[i + 1] : INT_MAX;
+ int Bleft = j >= 0 ? B[j] : INT_MIN;
+ int Bright = (j + 1) < BSize ? B[j + 1] : INT_MAX;
+
+ // partition is correct
+ if(Aleft <= Bright && Bleft <= Aright)
+ {
+ // Odd
+ if(total % 2)
+ {
+ return min(Aright, Bright);
+ }
+
+ // Even
+ return (max(Aleft, Bleft) + min(Aright, Bright)) / 2.0;
+ }
+ else if(Aleft > Bright)
+ {
+ r = i - 1;
+ }
+ else
+ {
+ l = i + 1;
+ }
+ }
+
+}