Find a number in sorted multidimentional array with binary search

Question

we got an increasing sorted multidimensional array for example:

int[][] mat = {{1,2,3,4},
{5,6,7,8},
{9,10,11,12},
{13,14,15,16}};

How can

Answer 1

public boolean Find(int[][] array, int number) { 
    int find = -1;
    for(int i = 0; i < N; i++) {
        find = binarySearch(array[i], number, 0, N);
        if(find != -1) { 
           return true; //the element is exist
        }
     }
     return false;//the element is not exist
}

Or you can revise this question it will help you a lot

Answer 2

2-d array can be used as 1-d array in following way using following formula for index:-

Say you need to find kth index in 1-d array then it be element with i= k/n and j = k%n where n is order of the matrix in the 2-d array. Use binary search as in 1-d array with ending index n*n-1.

Alternative Approach:-

1.> Do binary search on first elements of each 1-D array that arr[i][0].

2.> Then using above get the 1-D array which contains the element say k then do binary search on arr[k].

Answer 3

There are different ways to define 'order' in a two-dimensional array. For the order in your example, do bin search on the first element of each row, to find the row, and then do bin search again inside the row. if the number of rows is >= than the number of columns, you can optimize doing binary search on the elements of the diagonal that starts at (0,0), then do another binary search on the rest of the row.

Answer 4

You can use Arrays.binarySearch() for each sub array.

private static int[] binarySearch2d(int[][] arr, int toFind) {
    int[] index2d = new int[] { -1, -1 };

    // find the row
    int row = -1;
    for (int i = 0; i < arr.length; i++) {
        if (arr[i][0] > toFind) {
            break;
        }
        row = i;
    }

    if (row > -1) {
        int indexInSecond = Arrays.binarySearch(arr[row], toFind);
        if (indexInSecond > -1) {
            index2d[0] = row;
            index2d[1] = indexInSecond;
        }
    }
    return index2d;
}

private static void test() {
    int[][] mat = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 },
            { 13, 14, 15, 16 } };

    int[] found = binarySearch2d(mat, 12);
    int element = mat[found[0]][found[1]];
    System.out.println("Found: " + element + " at mat[" + found[0] + "]["
            + found[1] + "]");
}

will output

Found: 12 at mat[2][3]

Answer 5

You could do a binary search using just the first element of each inner array to find which row it is in. Then do a binary search inside that row to find the column.

You might need a little more logic if your matrix supports duplicates.

Answer 6

Use java.util.Arrays. Use Arrays.binarySearch() function on flattened matrix:

int[][] mat = {{1,2,3,4},
{5,6,7,8},
{9,10,11,12},
{13,14,15,16}};


int key = 3;
int[] oneDArray = new int[mat[0].length*mat[0].length];

int s = 0;
for(int i = 0; i < mat[0].length; i ++) 
    for(int j = 0; j < mat[0].length; j ++){                           
        oneDArray[s] = mat[i][j];
        s++;
    } 


int found = Arrays.binarySearch(oneDArray, key);
if(found > -1){
     System.out.println(found/ mat[0].length + "," + found % mat[0].length);    
}

And the demo: https://ideone.com/bFZVMs