I had to write a code (as an exercise) that receives a 2D (square) row wise and col wise sorted array and an element, and return true is the element exists in the array.
The first thing that came to mind when i heard "sorted" is binary search, but than i realized that the last element in each row isn't necessarily smaller than the first one in the next line.
So, i figured out that the best complexity will be O(n), and wrote the following code:
public static boolean findN(int[][] a, int x) {
if (a.length == 0 || a[0].length == 0 || x > a[a.length - 1][a[0].length - 1] || x < a[0][0]) {
return false;
}
int LastRow = a.length - 1, Lastcol = a[0].length - 1, row = 0, col = 0;
while (row <= LastRow) {
if (a[row][col] == x) {
return true;
} else if (col < Lastcol) {
col++;
} else {
col = 0;
row++;
}
}
return false;
}
array example:
int [] [] arr = {{1,2,7,30}
{2,4,18,50}
{3,6,19,90}
{4,7,20,91}}
I came across a similar problem a few months ago and here is the code I found that works in O(logN + logM) [assuming the array is sorted row-wise as well as column-wise].
[...] but than i realized that the last element in each row isn't necessarily smaller than the first one in the next line. - In this case, you cannot achieve O(logn) complexity.
Simple Binary Search:
static void binarySearch(int mat[][], int i, int j_low, int j_high, int x) {
while (j_low <= j_high) {
int j_mid = (j_low + j_high) / 2;
// Element found
if (mat[i][j_mid] == x) {
System.out.println ( "Found at (" + i + ", " + j_mid +")");
return;
}
else if (mat[i][j_mid] > x)
j_high = j_mid - 1;
else
j_low = j_mid + 1;
}
System.out.println ( "Element no found");
}
Core Logic:
static void sortedMatrixSearch(int mat[][], int n, int m, int x) {
// Single row matrix
if (n == 1) {
binarySearch(mat, 0, 0, m - 1, x);
return;
}
// Do binary search in middle column.
// Condition to terminate the loop when the
// 2 desired rows are found
int i_low = 0;
int i_high = n - 1;
int j_mid = m / 2;
while ((i_low + 1) < i_high) {
int i_mid = (i_low + i_high) / 2;
// element found
if (mat[i_mid][j_mid] == x) {
System.out.println ( "Found at (" + i_mid +", " + j_mid +")");
return;
}
else if (mat[i_mid][j_mid] > x)
i_high = i_mid;
else
i_low = i_mid;
}
// If element is present on
// the mid of the two rows
if (mat[i_low][j_mid] == x)
System.out.println ( "Found at (" + i_low + "," + j_mid +")");
else if (mat[i_low + 1][j_mid] == x)
System.out.println ( "Found at (" + (i_low + 1) + ", " + j_mid +")");
// Ssearch element on 1st half of 1st row
else if (x <= mat[i_low][j_mid - 1])
binarySearch(mat, i_low, 0, j_mid - 1, x);
// Search element on 2nd half of 1st row
else if (x >= mat[i_low][j_mid + 1] && x <= mat[i_low][m - 1])
binarySearch(mat, i_low, j_mid + 1, m - 1, x);
// Search element on 1st half of 2nd row
else if (x <= mat[i_low + 1][j_mid - 1])
binarySearch(mat, i_low + 1, 0, j_mid - 1, x);
// search element on 2nd half of 2nd row
else
binarySearch(mat, i_low + 1, j_mid + 1, m - 1, x);
}
Driver method:
public static void main (String[] args) {
int n = 4, m = 5, x = 8;
int mat[][] = {{0, 6, 8, 9, 11},
{20, 22, 28, 29, 31},
{36, 38, 50, 61, 63},
{64, 66, 100, 122, 128}};
sortedMatrixSearch(mat, n, m, x);
}
Hope this helps. Good luck.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.