Given an array of 0 and 1, find minimum no. of swaps to bring all 1s together (only adjacent swaps allowed)

Question

If given an array of 1\'s and 0\'s, what\'s good algorithm to show the minimum number of adjacent swaps needed to group all of the 1\'s together. The 1\'s don\'t need to be

Answer 1

Here is a simple, but not very clever algorithm that will perform an exhaustive search for any input in the range [0, 255].

• Input:
  • binary string
• Output:
  • optimal number of steps
  • number of optimal solutions
  • one detailed example

var transition = [],
    isSolution = [];

function init() {
  var msk = [ 3, 6, 12, 24, 48, 96, 192 ],
      i, j, n, x, cnt, lsb, msb, sz = [];

  for(i = 0; i < 0x100; i++) {
    for(n = cnt = msb = 0, lsb = 8; n < 8; n++) {
      if(i & (1 << n)) {
        cnt++;
        lsb = Math.min(lsb, n);
        msb = Math.max(msb, n);
      }
    }
    sz[i] = msb - lsb;
    isSolution[i] = (sz[i] == cnt - 1);
  }
  for(i = 0; i < 0x100; i++) {
    for(j = 0, transition[i] = []; j < 0x100; j++) {
      x = i ^ j;
      if(msk.indexOf(x) != -1 && (x & i) != x && (x & j) != x && sz[j] <= sz[i]) {
        transition[i].push(j);
      }
    }
  }
}

function solve() {
  var x = parseInt(document.getElementById('bin').value, 2),
      path = [ x ],
      list = [],
      i, min, sol = [], res = [];

  recurse(x, path, list);

  for(i in list) {
    if(min === undefined || list[i].length <= min) {
      min = list[i].length;
      (sol[min] = (sol[min] || [])).push(list[i]);
    }
  }
  console.log('Optimal length: ' + (min - 1) + ' step(s)');
  console.log('Number of optimal solutions: ' + sol[min].length);
  console.log('Example:');

  for(i in sol[min][0]) {
    res.push(('0000000' + sol[min][0][i].toString(2)).substr(-8, 8));
  }
  console.log(res.join(' -> '));
}

function recurse(x, path, list) {
  if(isSolution[x]) {
    list.push(path);
    return;
  }
  for(i in transition[x]) {
    if(path.indexOf(y = transition[x][i]) == -1) {
      recurse(y, path.slice().concat(y), list);
    }
  }
}

init();

solve