Minimum number of swaps to convert a string into another string [duplicate] - c

Question

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Finding the minimum number of swaps to convert one string to another, where the strings may have repeated characters
(3 answers)
Closed 8 years ago.
Problem statement:
Find the minimum number of swaps to convert one string into another of the same length, which may or may not have duplicate characters; arbitrary swaps are allowed.
min_swaps('kamal', 'amalk') -> 3
# 1 2 3
# kamal -> lamak -> aamlk -> amalk
Note: There are many of these questions on SO, but none that seem to apply to arbitrary swaps.
Initial Approach:
let s1 = 'kamal'
let s2 = 'amalk'
Assume that s1 is the "correct" ordering, that is its elements map to sequence from 0 -> N-1 in increasing order.
0 1 2 3 4
k a m a l
Now create an array P that is a mapping from letters in s2 to the correct index in s1:
1 2 3 4 0
a m a l k
P = [1,2,3,4,0]
Now we can count the number of array inversions in P using a modified mergesort, which will give us the number of elements that are out of order.
Modified mergesort:
int main(int argc, char ** argv) {
int array[] = { 1,2,3,4,0 };
int array_size = sizeof(array)/sizeof(array[0]);
int inversions = merge_sort(array, 0, array_size - 1);
printf("Found %d inversions\n", inversions);
return 0;
}
int merge_sort(int a[], int start, int end) {
if ( end > start ) {
int mid = start + (end - start) / 2;
int x = merge_sort(a, start, mid);
int y = merge_sort(a, mid + 1, end);
int z = merge(a, start, mid, end);
return x + y + z;
}
return 0;
}
int merge(int a[], int start, int mid, int end) {
int l = start, r = mid + 1;
int i = 0;
int temp[end - start + 1];
int splitInversionCount = 0;
while ( l <= mid && r <= end ) {
if ( a[l] < a[r] ) {
temp[i++] = a[l++];
}
else {
splitInversionCount += mid - l + 1;
temp[i++] = a[r++];
}
}
// copy whichever half of the array remains
while ( l <= mid ) {
temp[i++] = a[l++];
}
while ( r <= end ) {
temp[i++] = a[r++];
}
// copy temp back into a
int k;
for(k = 0; k < i; k++) {
a[k + start] = temp[k];
}
return splitInversionCount;
}
The problem with this is that it gives the minimum number of swaps with only adjacent elements, this returns 4 instead of 3.
Question:
Is there any way to extend this algorithm to capture arbitrary swaps as well? Or do I need a completely different approach?

Answer 1

I have not a full solution yet, but I think it's useful to state this to help others.
Let's assume the strings have size N. Let's call k the number of characters that are already in their position. Initially k <= N
A swap can have either:
Set 2 characters in the right position. Increase k by 2
Set 1 character in the right position. Increase k by 1
Do nothing useful, k remains the same.
You can always make a move of the second kind, and you can find it in O(n) time. Just take one character out of pos, and find the position where it needs to be, then swap.
You shouldn't make a swap of the first type as soon as you see one, as you could break too other moves that set 2 characters at the same time. If you do so you'll end up setting 4 chars in 3 moves when you could have done it in 2.
Making a move of the third kind may allow making 2 moves of the first kind, so all three kind of moves are useful.