double findMedianSortedArrays(int A[], int m, int B[], int n) {
if ( m==0 && n==0 ) return 0.0;
if ( m==0 ) return n%2==1 ? B[n/2] : (B[n/2-1] + B[n/2])/2.0;
if ( n==0 ) return m%2==1 ? A[m/2] : (A[m/2-1] + A[m/2])/2.0;
if ( m > n ){
return findMedianSortedArrayHelper(A, m, B, n, 0, m-1, 0, n-1);
}
return findMedianSortedArrayHelper(B, n, A, m, 0, n-1, 0, m-1);
}
double findMedianSortedArrays(int A[], int m, int B[], int n) {
//checking the edge cases
if ( m==0 && n==0 ) return 0.0;
//if the length of array is odd, return the middle one
//if the length of array is even, return the average of the middle two numbers
if ( m==0 ) return n%2==1 ? B[n/2] : (B[n/2-1] + B[n/2])/2.0;
if ( n==0 ) return m%2==1 ? A[m/2] : (A[m/2-1] + A[m/2])/2.0;
//let the longer array be A, and the shoter array be B
if ( m > n ){
return findMedianSortedArrayHelper(A, m, B, n, 0, m-1, 0, n-1);
}
return findMedianSortedArrayHelper(B, n, A, m, 0, n-1, 0, m-1);
}
double findMedianSortedArrayHelper(int A[], int m, int B[], int n, int lowA, int highA, int lowB, int highB) {
int mid = lowA + (highA - lowA)/2;
int pos = binarySearch(B, lowB, highB, A[mid]);
int num = mid + pos;
if (num == (m+n)/2){
if ((m+n)%2==1){
return A[mid];
}
int next;
if (mid>0 && pos>0){
next = A[mid-1]>B[pos-1] ? A[mid-1] : B[pos-1];
}else if(pos>0){
next = B[pos-1];
}else if(mid>0){
next = A[mid-1];
}
return (A[mid] + next)/2.0;
}
if (num < (m+n)/2){
lowA = mid + 1;
lowB = pos;
if ( highA - lowA > highB - lowB ) {
return findMedianSortedArrayHelper(A, m, B, n, lowA, highA, lowB, highB);
}
return findMedianSortedArrayHelper(B, n, A, m, lowB, highB, lowA, highA);
}
if (num > (m+n)/2) {
highA = mid - 1;
highB = pos-1;
if ( highA - lowA > highB - lowB ) {
return findMedianSortedArrayHelper(A, m, B, n, lowA, highA, lowB, highB);
}
return findMedianSortedArrayHelper(B, n, A, m, lowB, highB, lowA, highA);
}
}
double findMedianSortedArrayHelper(int A[], int m, int B[], int n, int lowA, int highA, int lowB, int highB) {
// Take the A[middle], search its position in B array
int mid = lowA + (highA - lowA)/2;
int pos = binarySearch(B, lowB, highB, A[mid]);
int num = mid + pos;
// If the A[middle] in B is B's middle place, then we can have the result
if (num == (m+n)/2){
// If two arrays total length is odd, just simply return the A[mid]
// Why not return the B[pos] instead ?
// suppose A={ 1,3,5 } B={ 2,4 }, then mid=1, pos=1
// suppose A={ 3,5 } B={1,2,4}, then mid=0, pos=2
// suppose A={ 1,3,4,5 } B={2}, then mid=1, pos=1
// You can see, the `pos` is the place A[mid] can be inserted, so return A[mid]
if ((m+n)%2==1){
return A[mid];
}
// If tow arrys total length is even, then we have to find the next one.
int next;
// If both `mid` and `pos` are not the first postion.
// Then, find max(A[mid-1], B[pos-1]).
// Because the `mid` is the second middle number, we need to find the first middle number
// Be careful about the edge case
if (mid>0 && pos>0){
next = A[mid-1]>B[pos-1] ? A[mid-1] : B[pos-1];
}else if(pos>0){
next = B[pos-1];
}else if(mid>0){
next = A[mid-1];
}
return (A[mid] + next)/2.0;
}
// if A[mid] is in the left middle place of the whole two arrays
//
// A(len=16) B(len=10)
// [................] [...........]
// ^ ^
// mid=7 pos=1
//
// move the `low` pointer to the "middle" position, do next iteration.
if (num < (m+n)/2){
lowA = mid + 1;
lowB = pos;
if ( highA - lowA > highB - lowB ) {
return findMedianSortedArrayHelper(A, m, B, n, lowA, highA, lowB, highB);
}
return findMedianSortedArrayHelper(B, n, A, m, lowB, highB, lowA, highA);
}
// if A[mid] is in the right middle place of the whole two arrays
if (num > (m+n)/2) {
highA = mid - 1;
highB = pos-1;
if ( highA - lowA > highB - lowB ) {
return findMedianSortedArrayHelper(A, m, B, n, lowA, highA, lowB, highB);
}
return findMedianSortedArrayHelper(B, n, A, m, lowB, highB, lowA, highA);
}
}
