Search element in a Spirally sorted Matrix

Given a spirally sorted matrix with N * N elements and an integer X, the task is to find the position of this given integer in the matrix if it exists, else print -1. Note that all the matrix elements are distinct.

Examples:

Input: arr[] = {
{1, 2, 3, 4},
{12, 13, 14, 5},
{11, 16, 15, 6},
{10, 9, 8, 7}}, X = 9
Output: 3 1
9 appears in row number 3 and column number 1 (0-based indexing)
Thus, output is (3, 1).

Input: arr[] = {
{1, 2, 3},
{8, 9, 4},
{7, 6, 5}}, X = 9
Output: 1 1

A simple solution is to search through all the elements in the array. The worst case time complexity of this approach will be O(n2).

A better solution is to use binary search. We apply binary search in two phases.
But before jumping to that, lets define what a ring means in here. A ring is defined as a set of all the the cells in the array such that there minimum of the distances from all the four sides is equal.

First, we try to determine the ring the number ‘X’ will belong to. We will do this using binary search. For that, observe the diagonal elements of the matrix. First ceil(N/2) of the diagonal matrix are guaranteed to be sorted in increasing order. So, each one of the ceil(N/2) diagonal elements can represent a ring. By, applying binary on the first ceil(N/2) diagonal elements, we determine the ring the number ‘X’ belongs to in O(log(n)) time.
After that, we apply binary search on the elements of the ring. Before that we determine the side of the ring, the number ‘X’ will belong to. Then, we apply the binary search correspondingly.

So, the total time complexity becomes O(log(n)).

Below is the implementation of the above approach:

C++

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// C++ implementation of the above approach
  
#include <iostream>
#define n 4
using namespace std;
  
// Function to return the ring, the number x
// belongs to.
int findRing(int arr[][n], int x)
{
    // Returns -1 if number x is smaller than
    // least element of arr
    if (arr[0][0] > x)
        return -1;
  
    // l and r represent the diagonal
    // elements to search in
    int l = 0, r = (n + 1) / 2 - 1;
  
    // Returns -1 if number x is greater
    // than the largest element of arr
    if (n % 2 == 1 && arr[r][r] < x)
        return -1;
    if (n % 2 == 0 && arr[r + 1][r] < x)
        return -1;
  
    while (l < r) {
        int mid = (l + r) / 2;
        if (arr[mid][mid] <= x)
            if (mid == (n + 1) / 2 - 1
                || arr[mid + 1][mid + 1] > x)
                return mid;
            else
                l = mid + 1;
        else
            r = mid - 1;
    }
      
    return r;
}
  
// Function to perform binary search 
// on an array sorted in increasing order
// l and r represent the left and right 
// index of the row to be searched
int binarySearchRowInc(int arr[][n], int row,
                    int l, int r, int x)
{
    while (l <= r) {
        int mid = (l + r) / 2;
          
        if (arr[row][mid] == x)
            return mid;
              
        if (arr[row][mid] < x)
            l = mid + 1;
        else
            r = mid - 1;
    }
      
    return -1;
}
  
// Function to perform binary search on 
// a particlar column of the 2D array
// t and b represent top and 
// bottom rows
int binarySearchColumnInc(int arr[][n], int col,
                        int t, int b, int x)
{
    while (t <= b) {
          
        int mid = (t + b) / 2;
          
        if (arr[mid][col] == x)
            return mid;
          
        if (arr[mid][col] < x)
            t = mid + 1;
        else
            b = mid - 1;
    }
      
    return -1;
}
  
// Function to perform binary search on 
// an array sorted in decreasing order
int binarySearchRowDec(int arr[][n], int row,
                    int l, int r, int x)
{
    while (l <= r) {
          
        int mid = (l + r) / 2;
          
        if (arr[row][mid] == x)
            return mid;
          
        if (arr[row][mid] < x)
            r = mid - 1;
        else
            l = mid + 1;
    }
      
    return -1;
}
  
// Function to perform binary search on a
// particlar column of the 2D array
int binarySearchColumnDec(int arr[][n], int col,
                        int t, int b, int x)
{
    while (t <= b) {
        int mid = (t + b) / 2;
          
        if (arr[mid][col] == x)
            return mid;
          
        if (arr[mid][col] < x)
            b = mid - 1;
        else
            t = mid + 1;
    }
      
    return -1;
}
  
// Function to find the position of the number x
void spiralBinary(int arr[][n], int x)
{
  
    // Finding the ring
    int f1 = findRing(arr, x);
  
    // To store row and column
    int r, c;
  
    if (f1 == -1) {
        cout << "-1";
        return;
    }
  
    // Edge case if n is odd
    if (n % 2 == 1 && f1 == (n + 1) / 2 - 1) {
        cout << f1 << " " << f1 << endl;
        return;
    }
  
    // Check which of the 4 sides, the number x
    // lies in
    if (x < arr[f1][n - f1 - 1]) {
        c = binarySearchRowInc(arr, f1, f1,
                            n - f1 - 2, x);
        r = f1;
    }
    else if (x < arr[n - f1 - 1][n - f1 - 1]) {
        c = n - f1 - 1;
          
        r = binarySearchColumnInc(arr, n - f1 - 1, f1,
                                n - f1 - 2, x);
    }
    else if (x < arr[n - f1 - 1][f1]) {
          
        c = binarySearchRowDec(arr, n - f1 - 1, f1 + 1,
                            n - f1 - 1, x);
        r = n - f1 - 1;
    }
    else {
          
        r = binarySearchColumnDec(arr, f1, f1 + 1,
                                n - f1 - 1, x);
        c = f1;
    }
  
    // Printing the position
    if (c == -1 || r == -1)
        cout << "-1";
    else
        cout << r << " " << c;
  
    return;
}
  
// Driver code
int main()
{
    int arr[][n] = { { 1, 2, 3, 4 },
                    { 12, 13, 14, 5 },
                    { 11, 16, 15, 6 },
                    { 10, 9, 8, 7 } };
  
    spiralBinary(arr, 7);
  
    return 0;
}

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Java

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// Java implementation of the above approach 
class GFG 
{
  
final static int n =4
  
// Function to return the ring, 
// the number x belongs to. 
static int findRing(int arr[][], int x) 
    // Returns -1 if number x is  
    // smaller than least element of arr 
    if (arr[0][0] > x) 
        return -1
  
    // l and r represent the diagonal 
    // elements to search in 
    int l = 0, r = (n + 1) / 2 - 1
  
    // Returns -1 if number x is greater 
    // than the largest element of arr 
    if (n % 2 == 1 && arr[r][r] < x) 
        return -1
    if (n % 2 == 0 && arr[r + 1][r] < x) 
        return -1
  
    while (l < r) 
    
        int mid = (l + r) / 2
        if (arr[mid][mid] <= x) 
            if (mid == (n + 1) / 2 - 1
                || arr[mid + 1][mid + 1] > x) 
                return mid; 
            else
                l = mid + 1
        else
            r = mid - 1
    
    return r; 
  
// Function to perform binary search 
// on an array sorted in increasing order 
// l and r represent the left and right 
// index of the row to be searched 
static int binarySearchRowInc(int arr[][], int row, 
                    int l, int r, int x) 
    while (l <= r) 
    
        int mid = (l + r) / 2
          
        if (arr[row][mid] == x) 
            return mid; 
              
        if (arr[row][mid] < x) 
            l = mid + 1
        else
            r = mid - 1
    
    return -1
  
// Function to perform binary search on 
// a particlar column of the 2D array 
// t and b represent top and 
// bottom rows 
static int binarySearchColumnInc(int arr[][], int col, 
                        int t, int b, int x) 
    while (t <= b)
    
        int mid = (t + b) / 2
          
        if (arr[mid][col] == x) 
            return mid; 
          
        if (arr[mid][col] < x) 
            t = mid + 1
        else
            b = mid - 1
    
    return -1
  
// Function to perform binary search on 
// an array sorted in decreasing order 
static int binarySearchRowDec(int arr[][], int row, 
                    int l, int r, int x) 
    while (l <= r) { 
          
        int mid = (l + r) / 2
          
        if (arr[row][mid] == x) 
            return mid; 
          
        if (arr[row][mid] < x) 
            r = mid - 1
        else
            l = mid + 1
    
    return -1
  
// Function to perform binary search on a 
// particlar column of the 2D array 
static int binarySearchColumnDec(int arr[][], int col, 
                        int t, int b, int x) 
    while (t <= b) 
    
        int mid = (t + b) / 2
          
        if (arr[mid][col] == x) 
            return mid; 
          
        if (arr[mid][col] < x) 
            b = mid - 1
        else
            t = mid + 1
    
    return -1
  
// Function to find the position of the number x 
static void spiralBinary(int arr[][], int x) 
  
    // Finding the ring 
    int f1 = findRing(arr, x); 
  
    // To store row and column 
    int r, c; 
  
    if (f1 == -1
    
            System.out.print("-1");
        return
    
  
    // Edge case if n is odd 
    if (n % 2 == 1 && f1 == (n + 1) / 2 - 1)
    
            System.out.println(f1+" "+f1);
        return
    
  
    // Check which of the 4 sides, the number x 
    // lies in 
    if (x < arr[f1][n - f1 - 1])
    
        c = binarySearchRowInc(arr, f1, f1, 
                            n - f1 - 2, x); 
        r = f1; 
    
    else if (x < arr[n - f1 - 1][n - f1 - 1]) 
    
        c = n - f1 - 1
          
        r = binarySearchColumnInc(arr, n - f1 - 1, f1, 
                                n - f1 - 2, x); 
    
    else if (x < arr[n - f1 - 1][f1])
    
        c = binarySearchRowDec(arr, n - f1 - 1, f1 + 1
                            n - f1 - 1, x); 
        r = n - f1 - 1
    
    else
    
        r = binarySearchColumnDec(arr, f1, f1 + 1
                                n - f1 - 1, x); 
        c = f1; 
    
  
    // Printing the position 
    if (c == -1 || r == -1
        System.out.print("-1"); 
    else
            System.out.print(r+" "+c);
  
    return
  
// Driver code 
public static void main(String[] args) 
{
        int arr[][] = { { 1, 2, 3, 4 }, 
                    { 12, 13, 14, 5 }, 
                    { 11, 16, 15, 6 }, 
                    { 10, 9, 8, 7 } }; 
  
    spiralBinary(arr, 7); 
}
}
  
// This code is contributed by 29AjayKumar

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Output:

3 3


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