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

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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