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Thursday Code Puzzler: Matrix Search

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Thursday is code puzzler day here at DZone. The idea is simple: solve the coding problem as efficiently as you can, in any language or framework that you find suitable.

Note: Even though there really is nothing stopping you from finding a solution to this on the internet, try to keep honest, and come up with your own answer.  It's all about the participation!

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Matrix Search 

Given a matrix of 1s and 0s, write an algorithm that will find the groups of 1s. A group is a series of 1's that are connected in any direction. An example matrix would look as follows, defined as a 2d array of integers


Catch up on all our previous puzzlers here.



Andrew Gilmartin replied on Thu, 2013/05/02 - 7:21am

Can the group have holes? For example, is this

1 1 1
1 0 1
0 1 0
one group, ie 
a a a
a   a
or two, ie
a a a
a   a
-- Andrew

Niranjan Tallapalli replied on Thu, 2013/05/02 - 9:07am

There might be better time complexity than what I have written but here is my approach, I have also blogged this solution on my blog 

int findMaxDepth(int[][] n, boolean[][] bool, int i, int j) {
    // Check if i, j (indexes) are within the size of array
    // Check if the cell is already traversed or not using bool array
    // Check if the cell value is 1 before counting it as part of a sector
    if(i >= 0 && i < n.length &&                  j >=0 && j < n[0].length
                && bool[i][j] != true && n[i][j] != 0) {
        // Mark the status of the cell for backtracking purpose
        bool[j] = true;
        // left traversal
        findMaxDepth(n, bool, i-1, j);
        // right traversal
        findMaxDepth(n, bool, i+1, j);
        // top traversal
        findMaxDepth(n, bool, i, j-1);
        // bottom traversal
        findMaxDepth(n, bool, i, j+1);
        // Top-Bottom diagnol
        // diagnol-down traversal
        findMaxDepth(n, bool, i+1, j+1);
        // diagnol-up traversal
        findMaxDepth(n, bool, i-1, j-1);
        // Bottom-Top diagnol
        // diagnol-down traversal
        findMaxDepth(n, bool, i+1, j-1);
        // diagnol-up traversal
        findMaxDepth(n, bool, i-1, j+1);
    return currentDepth;

Ben Ritchie replied on Thu, 2013/05/09 - 7:18pm

// our provided matrix
var matrix[int][int]

// Define a new matrix rowstripsize[int][int] is either 0 or an int representing number of horizontal neighbours - on this row or above - connected to this matrix index i,j containing a 1
var rowstripsize[int][int]

computeGroups(matrix) {

  // populate rowstripsize
  for (int i = 0; i<matrix.height-1; i++) {

  maxGroupSize = 0

  // easily build reports on groups
  for (cell : rowstripsize) {
    maxGroupSize = max(maxGroupSize, cell)

  print "max group size is " + maxGroupSize

// populate rowstripsizes for a row
populateRowSizes(int i) {

  var seriesOfOnes = 0

  for(int j =0; j< matrix.width-1; j++) {
    if (matrix[i][j] = 1) { // another one - possibly first
        seriesOfOnes ++  
    } else if (seriesOfOnes > 0) { // end of a run of 1s - fill in stripsize in rowstripsize
       fillInStripSize(i, j, seriesOfOnes)
       // reset counter
       seriesOfOnes = 0
    } else { 
      // do nothing its a lone zero

// update rowstripsize for this strip considering stripsizes above
fillInStripSize(int i, int finalOne, int seriesOfOnes) {

  // if cells above our strip have a strip size then we should add it to our group total
  // in large matrices there may be muliple strips above our strip! So only count new strips
  var stripAboveCounted = false 
  // our grand total for this strip so far
  var totalGroupSizeSoFar = seriesOfOnes

  // Add all unique stripsizes above, including those connected by diaganols
  // So define the start and end index..

  var aboveStartSearchIndex = finalOne - seriesOfOnes;
  // allow for diagonols by trying going back one in above strip search
  aboveStartSearchIndex = max(0, aboveStartSearchIndex - 1);

  // allow for diagonols by trying going fwd one in above strip search
  var aboveEndSearchIndex = min(matrix.width, finalOne + 1);

  // Now we can add up the unique stripssizes connected to this strip above
  for (int j=aboveStartSearchIndex; j <= aboveEndSearchIndex; j++) {
      var stripSizeAbove = rowstripsize[i-1][j]

      if (stripSizeAbove > 0 && !stripAboveCounted) { 
        totalGroupSizeSoFar += stripSizeAbove
        stripAboveCounted = true
      } else if (stripSizeAbove = 0 && stripAboveCounted) {
        stripAboveCounted = false

  // Now we know the grand total for this strip - populate it
  for (int j=finalOne - seriesOfOnes; j <= finalOne; j++) {
    rowstripsize[i][j] = totalGroupSizeSoFar

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