[Leetcode][240. Search a 2D Matrix II] 5 Approaches: BF, Binary Search(Row), Binary Search(Diagonal, Row, Col), Binary Search(Global), 2D Coord Axis

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By Long Luo

This article is the solution 5 Approaches: BF, Binary Search(Row), Binary Search(Diagonal, Row, Col), Binary Search(Global), 2D Coord Axis of Problem 240. Search a 2D Matrix II.

The similar question and solution is 74. Search a 2D Matrix.

Here shows 5 Approaches to slove this problem: Brute Force, Binary Search(Row), Binary Search(Diagonal, Row, Col), Binary Search(Global), 2D Coord Axis.

Brute Force

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public static boolean searchMatrix_bf(int[][] matrix, int target) {
    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
        return false;
    }

    int row = matrix.length;
    int col = matrix[0].length;

    if (matrix[0][0] > target || matrix[row - 1][col - 1] < target) {
        return false;
    }

    for (int i = 0; i < row; i++) {
        if (matrix[i][col - 1] < target) {
            continue;
        }

        for (int j = 0; j < col; j++) {
            if (matrix[i][j] == target) {
                return true;
            }
        }
    }

    return false;
}

Analysis

  • Time Complexity: O(mn)O(mn).
  • Space Complexity: O(1)O(1).

Binary Search(Row)

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public boolean searchMatrix_bs_row(int[][] matrix, int target) {
    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
        return false;
    }

    int row = matrix.length;
    int col = matrix[0].length;

    if (matrix[0][0] > target || matrix[row - 1][col - 1] < target) {
        return false;
    }

    for (int i = 0; i < row; i++) {
        if (matrix[i][col - 1] < target) {
            continue;
        }

        if (binarySearchRow(matrix[i], target)) {
            return true;
        }
    }

    return false;
}

public boolean binarySearchRow(int[] arr, int target) {
    int low = 0;
    int high = arr.length - 1;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (arr[mid] == target) {
            return true;
        } else if (arr[mid] > target) {
            high = mid - 1;
        } else if (arr[mid] < target) {
            low = mid + 1;
        }
    }

    return false;
}

Analysis

  • Time Complexity: O(min(M,N)(logM+logN))O(min(M,N)(logM+logN)).
  • Space Complexity: O(1)O(1).

Binary Search(Diagonal, Row, Col)

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public static boolean searchMatrix_bs_3d(int[][] matrix, int target) {
    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
        return false;
    }

    int row = matrix.length;
    int col = matrix[0].length;

    if (matrix[0][0] > target || matrix[row - 1][col - 1] < target) {
        return false;
    }

    int index = diagonalBinarySearch(matrix, target);
    if (matrix[index][index] == target) {
        return true;
    }

    for (int i = 0; i <= index; i++) {
        boolean rowResult = rowBinarySearch(matrix, i, col - 1, target);
        boolean colResult = colBinarySearch(matrix, i, row - 1, target);
        if (rowResult || colResult) {
            return true;
        }
    }

    return false;
}

public static int diagonalBinarySearch(int[][] matrix, int target) {
    int minVal = Math.min(matrix.length, matrix[0].length);
    int left = 0;
    int right = minVal;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (matrix[mid][mid] < target) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }

    return Math.min(left, minVal - 1);
}

public static boolean rowBinarySearch(int[][] matrix, int start, int end, int target) {
    int left = start;
    int right = end;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (matrix[start][mid] == target) {
            return true;
        } else if (matrix[start][mid] > target) {
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }

    return false;
}

public static boolean colBinarySearch(int[][] matrix, int start, int end, int target) {
    int left = start + 1;
    int right = end;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (matrix[mid][start] == target) {
            return true;
        } else if (matrix[mid][start] > target) {
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }

    return false;
}

Analysis

  • Time Complexity: O(min(M,N)(logM+logN))O(min(M,N)(logM+logN)).
  • Space Complexity: O(1)O(1).

Binary Search(Global)

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public static boolean searchMatrix_bs(int[][] matrix, int target) {
    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
        return false;
    }

    int row = matrix.length;
    int col = matrix[0].length;

    if (matrix[0][0] > target || matrix[row - 1][col - 1] < target) {
        return false;
    }

    return searchMatrixHelper(matrix, 0, 0, col - 1, row - 1, col - 1, row - 1, target);
}

private static boolean searchMatrixHelper(int[][] matrix, int x1, int y1, int x2, int y2, int xMax, int yMax, int target) {
    if (x1 > xMax || y1 > yMax) {
        return false;
    }

    if (x1 == x2 && y1 == y2) {
        return matrix[y1][x1] == target;
    }
    int m1 = (x1 + x2) >>> 1;
    int m2 = (y1 + y2) >>> 1;
    if (matrix[m2][m1] == target) {
        return true;
    }
    if (matrix[m2][m1] < target) {
        // Right Up
        return searchMatrixHelper(matrix, m1 + 1, y1, x2, m2, x2, y2, target) ||
                // Left Down
                searchMatrixHelper(matrix, x1, m2 + 1, m1, y2, x2, y2, target) ||
                // Right Down
                searchMatrixHelper(matrix, m1 + 1, m2 + 1, x2, y2, x2, y2, target);

    } else {
        // Right Up
        return searchMatrixHelper(matrix, m1 + 1, y1, x2, m2, x2, y2, target) ||
                // Left Down
                searchMatrixHelper(matrix, x1, m2 + 1, m1, y2, x2, y2, target) ||
                // Left Up
                searchMatrixHelper(matrix, x1, y1, m1, m2, x2, y2, target);
    }
}

Analysis

  • Time Complexity: O(mlogn)O(mlogn).
  • Space Complexity: O(1)O(1).

2D Coord Axis

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public static boolean searchMatrix_coord_left(int[][] matrix, int target) {
    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
        return false;
    }

    int row = matrix.length;
    int col = matrix[0].length;

    if (matrix[0][0] > target || matrix[row - 1][col - 1] < target) {
        return false;
    }

    int i = row - 1;
    int j = 0;
    while (i >= 0 && j < col) {
        if (target > matrix[i][j]) {
            j++;
        } else if (target < matrix[i][j]) {
            i--;
        } else {
            return true;
        }
    }

    return false;
}

Analysis

  • Time Complexity: O(m+n)O(m + n).
  • Space Complexity: O(1)O(1).

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