Given a binary string S, the task is to count the minimum number of jumps required to group all 1’s together.
Input: S = “000010011000100”
000010011000100 -> 000000111000100 requires 2 jumps.
000000111000100 -> 000000111100000 requires 3 jumps.
Hence, at least 5 jumps are required to group all 1’s together.
Input: S = “100010001”
100010001 -> 000110001 requires 3 jumps.
000110001 -> 000111000 requires 3 jumps.
We can observe that in order to minimize the number of jumps required for grouping all 1’s together, they need to be grouped near the median of their current positions. Calculate the median and the number of moves required to shift the 1’s to the nearest position of 0 in the left of the median. Perform the same operation for the right of the median.
Below is the implementation of the above approach:
Time Complexity: O(N)
Auxiliary Space: O(N)
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