Young Tableau

A a X b matrix is Young Tableau if all rows(from left to right) and columns(from top to bottom) are sorted. Some of the entries in matrix can be an INFINITY (MAX) value to represent that its not a valid value.

Example

1 5 9 4 7 11 8 12 MAX

Operations supported on Young Tableau

There are majorly three operations supported here:

1. Prepare Young Tableau from a given 2-d array or a matrix

Given an input matrix, you want to make it honor young tableau property.

2. Extract a minimum value (and delete it)

The minimum that resides on (0,0) position. Return it, and delete from matrix. We need to do some swap operations in order to make this matrix again as Young Tableau.

3. Search an element in young tableau

Given an element, we need to search in matrix and return its index in matrix.

Basic Building Pieces

Before we get started, lets understand some basics. This problem can be considered as a modified case of Min-Heap data structure. Where all elements below the root element are lesser. Here, we the the children of a matrix cell will be its right and bottom cells.

Some basic operations

/** * Return [-1,-1], if there is no child element */ private int [] getLeft ( int r , int c ) { if ( r >= this . size - 1 ) { return new int []{- 1 , - 1 }; } return new int []{ r + 1 , c }; } /** * Return [-1,-1], if there is no child element */ private int [] getRight ( int r , int c ) { if ( c >= this . size - 1 ) { return new int []{- 1 , - 1 }; } return new int []{ r , c + 1 }; } /** * Check if the cell is within the matrix */ private boolean isCellValid ( int [] cell ) { return cell [ 0 ]>= 0 && cell [ 0 ]< this . size && cell [ 1 ]>= 0 && cell [ 1 ]< this . size ; }

Min Heapify operation

private void tableau_heapify ( int r , int c ) { int [] left = this . getLeft ( r , c ); int [] right = this . getRight ( r , c ); int [] smallest = new int []{ r , c }; // System.out.println(r + ", " + c); //check if the cell is less than both of its children if ( this . isCellValid ( left ) && this . matrix [ left [ 0 ]][ left [ 1 ]] < this . matrix [ r ][ c ]) { //save address of left cell smallest = left ; } if ( this . isCellValid ( right ) && this . matrix [ right [ 0 ]][ right [ 1 ]] < this . matrix [ smallest [ 0 ]][ smallest [ 1 ]]) { //save address of right cell smallest = right ; } //Check if we found changed address in our smallest variables. //If yes, swap them. And, recursively call this method on that cell address if ( this . isCellValid ( smallest ) && ( smallest [ 0 ] != r || smallest [ 1 ] != c )) { //swap int t = this . matrix [ r ][ c ]; this . matrix [ r ][ c ] = this . matrix [ smallest [ 0 ]][ smallest [ 1 ]]; this . matrix [ smallest [ 0 ]][ smallest [ 1 ]] = t ; // recursive call tableau_heapify this . tableau_heapify ( smallest [ 0 ], smallest [ 1 ]); } }

It runs on an cell index, and check whether both children cells are lesser than or not. If not, swap with the minimum cell value, and keep on moving like this in recursion. Key points for heapify operation:

Get position of left and right children, and check if they are valid indexes (not crossing the matrix bounds)

Check, if any children is having a lesser value.

If found lesser value cell. Swap with it, and go in recursion.

Note, if both children are smaller. We will go with minimum value cell.

Extract Min

/** * Get element at (0,0), and make matrix tableau_heapify again */ public int extractMin () { int min = this . matrix [ 0 ][ 0 ]; //put a max value here, just to denote that it has been taken out. this . matrix [ 0 ][ 0 ] = Integer . MAX_VALUE ; //lets maintain tableau property this . tableau_heapify ( 0 , 0 ); //all is set return min ; }

Key points:

We know the min element is at (0,0) index

Get it, and replace it with an INFINITY(MAX) value.

Call our heapify() on this index.

return the min value.

Search an element

/** * a recursive method to search an element * @return null if not found */ private int [] search ( int element , int [] currentCell ) { if (! this . isCellValid ( currentCell )) { return null ; } int currentElement = this . matrix [ currentCell [ 0 ]][ currentCell [ 1 ]]; if ( element == currentElement ) { return currentCell ; } if ( element < currentElement ) { // move left // decrement col by 1 currentCell [ 1 ]--; return this . search ( element , currentCell ); } else { //(element > currentElement) { //move down //increment row by 0 currentCell [ 0 ] ++; return this . search ( element , currentCell ); } } /** * Search the given element and return its index in matrix */ public int [] searchElement ( int element ) { //we will start from top right corner to find the element return this . search ( element , new int []{ 0 , this . size - 1 }); }

Key points:

A simple brute force approach could be that keep on traversing from (0,0), and row wise. We can find our element somewhere in matrix. However, this will take O(mxn) runtime complexity.

An optimize approach can be if we can search either from left-bottom, or top-right, or bottom-right corner.

In this algorithm, we start from top-right.

There can be three cases: Either number found, just return it Number in current cell is greater. Simply goto its left. As, complete column is sorted. We cannot find it in this column. Number in current cell is lesser. Simply goto its bottom. As, complete row is sorted. Every element in left hand side will be lesser.

Do this in recursion, and either we find it, or not find it.

Image representation

Complete Code