Member Avatar for nataraja833

Hi Guys ,
I need guidence on how to approach this problem .

I have string of ones and zeros ( input ) and output has to be optimized following below rules
1) total number of one's have to minimum
2) zero from the string cannot be moved from their place
3) one can be swapped to x , denoting empty .
4) distance between two one cannot be more than 4 indices

test.png

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The thread’s examples from and the exchange with suggest the intended task is: given a fixed linear array where zeros cannot move and any original 1 may be turned into an empty mark x, choose the smallest set of original-1 positions to keep as 1 so that every original-1 position is within index-distance 4 (inclusive) of at least one kept 1. That interpretation removes ambiguity in rule 4 and explains why many original 1s become x in the posted outputs — kept 1s act as “centers” that cover nearby 1s within radius R=4.

This is the standard minimum-interval-covering / sensor-placement problem on a line and is solved optimally by a simple greedy sweep. Collect the sorted indices of all original 1s. Repeatedly take the leftmost uncovered 1, place a center at the rightmost 1 that is within R indices of it, then mark all 1s within R of that center as covered. Repeat until all original 1s are covered. Afterward, reconstruct the output string by leaving indices in the chosen center set as 1, turning other original 1s to x, and keeping zeros unchanged.

Example pseudocode:

R = 4
pos = sorted indices i where input[i] == 1
i = 0
kept = set()
while i < len(pos):
    s = pos[i]
    j = i
    while j+1 < len(pos) and pos[j+1] <= s + R:
        j += 1
    center = pos[j]
    kept.add(center)
    # skip all points covered by center
    i = j + 1
    while i < len(pos) and pos[i] <= center + R:
        i += 1
# reconstruct: positions in 'kept' -> '1'; other original 1s -> 'x'; zeros unchanged

Running time is O(m) where m is the number of original 1s (after extracting indices). Correctness: the greedy choice (rightmost center that still covers the leftmost uncovered point) maximizes forward coverage and is optimal for 1D fixed-radius covering. If rule 4 was meant differently (open/closed bound, or a constraint between kept-1 pairs instead of covering original 1s), the approach must be adapted; the above assumes the coverage interpretation that matches the posted outputs.

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Member Avatar for rproffitt

Two questions.

  1. How do you solve this now? The answer leads to you coding that process.
  2. Why did your question vanish at I wonder if you were asked to restate or expand.
Member Avatar for nataraja833

got it
tclsh solve.tcl
Input : 1 1 1 0 0 0 0 1 1 0 0 0 0 0 1 1 1 0 1 0 1 0
Output : 1 x 1 0 0 0 0 1 1 0 0 0 0 0 1 x x 0 1 0 x 0

Input : 1 1 1 0 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 1 1
Output : 1 x 1 0 0 0 0 1 0 0 0 0 0 1 x 1 0 0 0 1 0 x x x

Input : 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0
Output : 1 x x x x 1 x x x x 1 x x x 1 1 0 0 0 0 0 0 0 0 1 x x x x 1 1 0 0 0 0 0 0 0 0 0 0 0

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