Method findMax

largestLocal.java:12–31

(int[][] grid, int startRow, int startCol, PriorityQueue<Integer> pq)

  // Hence for every 3*3 matrix starting at the ith row and jth column ([i,j] cell and j>0) remove the (j-1)th column, add the (j+2)th column and find the maximum
  // at every new row, reinitialize the queue.
    public int findMax(int[][] grid, int startRow, int startCol, PriorityQueue<Integer> pq){
        //for every new row, init the queue and store 9 elements (3*3)
      // TC -> O(9log9) [for inserting 1 element in pq, tc is logK, where k is number of element, our pq will always have 9 elements at max]
        if(startCol==0){
            pq.clear();
            for(int i=startRow;i<startRow+3;i++){
                for(int j=startCol;j<startCol+3;j++){
                    pq.offer(grid[i][j]);
                }
            }
        }else{ 
          //TC -> 6log9 [remove -> 3log9 + insert 3log9]
          //for same row, use sliding window to remove the (j-1)th column and add (j+2) column
            for(int j=startRow;j<startRow+3;j++){
                pq.remove(grid[j][startCol-1]);
                pq.offer(grid[j][startCol+2]);
            }
        }
        return pq.peek();
    }
    public int[][] largestLocal(int[][] grid) {
        //priorityQueue to return the maximum values out of 9 values (3*3 matrix)
        PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());