Median of Two Sorted Arrays - With Detailed Explanation

LeetCode/Hard/4. Median of Two Sorted Arrays.cpp

@@ -13,6 +13,22 @@
 // Output: 2.50000
 // Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
 
+/*Suppose the two arrays are A and B.
+Perform the following variant of binary search first in A then B to find the median.
+
+Start from low = 0, high = |A|, guess i = floor (low + high)/2
+For the median m, there should be total half = floor (|A| + |B| + 1) / 2 elements not greater than it.
+Since there are i + 1 elements not greater than A[i] in A,
+There should be half - (i + 1) elements not greater than A[i] in B.
+Denote j = half - i - 2, thus we can compare if B[j] <= A[i] <= B[j + 1] is satisfied. This indicates
+That the guess is the correct median.
+
+Otherwise, we can easily tell if the guess is too small or too big, then halve the elements to adjust
+the guess.
+
+Fixes #134
+*/
+
 
 
 

LeetCode/Hard/median of two sorted arrays.cpp

@@ -0,0 +1,64 @@
+#define min(x, y) (x < y ? x : y)
+
+int odd(int n) { return n & 0x1; }
+
+void swap(int *x, int *y) {
+    int tmp = *x; *x = *y; *y = tmp;
+}
+
+/* median of an array */
+double medianof(int A[], int n) {
+    return odd(n) ? (double) A[n / 2] : (double)(A[ n / 2] + A[n / 2 - 1]) / 2.0;
+}
+
+int find(int A[], int m, int B[], int n) {
+    int l = 0, u = m;
+    int i, j, half = (m + n + 1) / 2;
+    if (!A || m == 0)
+        return medianof(B, n);
+    if (!B || n == 0)
+        return medianof(A, m);
+    while (l < u) {
+        i = (l + u) / 2;
+        j = half - i - 2;
+        if (j < 0 || j >= n) {
+            if (j == -1 && A[i] <= B[0])
+                return i; /* found */
+            if (j >= n )
+                l = i + 1; /* too small */
+            else
+                u = i; /* too big */
+        } else {
+            if (B[j]<= A[i] && (j == n - 1 || A[i] <= B[j+1]))
+                return i; /* found */
+            else if (A[i] < B[j])
+                l = i + 1; /* too small */
+            else
+                u = i; /* too big */
+        }
+    }
+    return -1;
+}
+
+double findMedianSortedArrays(int A[], int m, int B[], int n) {
+    int i, j, k, *C;
+    if (!A || m == 0)
+        return medianof(B, n);
+    if (!B || n == 0)
+        return medianof(A, m);
+    if ((i = find(A, m, B, n)) == -1) {
+        i = find(B, n, A, m);
+        C = A; A = B; B = C;
+        swap(&m, &n);
+    }
+    if (odd(m + n))
+        return (double)A[i];
+    j = (m + n) / 2 - i - 2;
+    if (i == m - 1)
+        k = B[j+1];
+    else if (j == n - 1)
+        k = A[i+1];
+    else
+        k = min(A[i+1], B[j+1]);
+    return (double)(A[i] + k) / 2.0;
+}