LeetCode – Search a 2D Matrix II (Java)

by

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

For example, consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Java Solution 1

In a naive approach, we can use the matrix boundary to reduce the search space. Here is a simple recursive implementation.

public boolean searchMatrix(int[][] matrix, int target) {
    int i1=0; 
    int i2=matrix.length-1;
    int j1=0;
    int j2=matrix[0].length-1;
 
    return helper(matrix, i1, i2, j1, j2, target);
}
 
public boolean helper(int[][] matrix, int i1, int i2, int j1, int j2, int target){
 
    if(i1>i2||j1>j2)
        return false;
 
    for(int j=j1;j<=j2;j++){
        if(target < matrix[i1][j]){
            return helper(matrix, i1, i2, j1, j-1, target);
        }else if(target == matrix[i1][j]){
            return true;
        }
    }
 
    for(int i=i1;i<=i2;i++){
        if(target < matrix[i][j1]){
            return helper(matrix, i1, i-1, j1, j2, target);
        }else if(target == matrix[i][j1]){
            return true;
        }
    }
 
    for(int j=j1;j<=j2;j++){
        if(target > matrix[i2][j]){
            return helper(matrix, i1, i2, j+1, j2, target);
        }else if(target == matrix[i2][j]){
            return true;
        }
    }
 
    for(int i=i1;i<=i2;i++){
        if(target > matrix[i][j2]){
            return helper(matrix, i1, i+1, j1, j2, target);
        }else if(target == matrix[i][j2]){
            return true;
        }
    }
 
    return false;
}

Java Solution 2

Time Complexity: O(m + n)

public boolean searchMatrix(int[][] matrix, int target) {
    int m=matrix.length-1;
    int n=matrix[0].length-1;
 
    int i=m; 
    int j=0;
 
    while(i>=0 && j<=n){
        if(target < matrix[i][j]){
            i--;
        }else if(target > matrix[i][j]){
            j++;
        }else{
            return true;
        }
    }
 
    return false;
}

3 thoughts on “LeetCode – Search a 2D Matrix II (Java)”

  1. there is a simple , logN solution

    find element X

    —

    step1

    start 0,0, lets find in first row elements bigger than X

    start 0.0 , lets find in first column element bigger than X

    Cross of row and column will give us point (m, n) somewhere in matrix, it is down/right element

    —

    step2

    from point (m,n), lets find elements in row M that smaller then X,

    from point (m,n), lets find elements in column N that smaller then X,
    cross of row and column will give us up/left element (F, G)

    ——-
    so we have sub matrix between (f,g) and (m,n)

    do over step1

  2. it is sorted and NO binary search?.. THERE MUST BE BETTER SOLUTION.!

    lets value =16

    by binary search in 0column, we can find smallest row, that bigger than value. so its row before 18.

    after for each column FROM 0 , check hi/low per column if value can be w/in,
    found 11-17 , 15-24, and can use binary search again.

  3. You don’t have condition for initial check.

    if(matrix == null || matrix.length == 0)
    {
    return false;
    }

    This should make Java solution 2 – acceptable in leetcode.

Leave a Comment