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, let’s define what a ring means here. A ring is defined as a set of all the cells in the array such that their minimum of the distances from all 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. The first ceil(N/2) of the diagonal matrix is 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 a 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++
// C++ implementation of the above approach#include <iostream>#define n 4using 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 searchedint 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 particular column of the 2D array// t and b represent top and // bottom rowsint 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 orderint 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// particular column of the 2D arrayint 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 xvoid 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 codeint 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;} |
Java
// 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 particular 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 // particular 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 |
Python3
# Python3 implementation of the above approach # Function to return the ring, # the number x belongs to. def findRing(arr, 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 l, r = 0, (n + 1) // 2 - 1 # Returns -1 if number x is greater # than the largest element of arr if n % 2 == 1 and arr[r][r] < x: return -1 if n % 2 == 0 and arr[r + 1][r] < x: return -1 while l < r: mid = (l + r) // 2 if arr[mid][mid] <= x: if (mid == (n + 1) // 2 - 1 or 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 def binarySearchRowInc(arr, row, l, r, x): while l <= r: mid = (l + r) // 2 if arr[row][mid] == x: return mid elif arr[row][mid] < x: l = mid + 1 else: r = mid - 1 return -1# Function to perform binary search on # a particular column of the 2D array # t and b represent top and # bottom rows def binarySearchColumnInc(arr, col, t, b, x): while t <= b: mid = (t + b) // 2 if arr[mid][col] == x: return mid elif 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 def binarySearchRowDec(arr, row, l, r, x): while l <= r: mid = (l + r) // 2 if arr[row][mid] == x: return mid elif arr[row][mid] < x: r = mid - 1 else: l = mid + 1 return -1# Function to perform binary search on a # particular column of the 2D array def binarySearchColumnDec(arr, col, t, b, x): while t <= b: mid = (t + b) // 2 if arr[mid][col] == x: return mid elif arr[mid][col] < x: b = mid - 1 else: t = mid + 1 return -1# Function to find the position of the number x def spiralBinary(arr, x): # Finding the ring f1 = findRing(arr, x) # To store row and column r, c = None, None if f1 == -1: print("-1") return # Edge case if n is odd if n % 2 == 1 and f1 == (n + 1) // 2 - 1: print(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 elif x < arr[n - f1 - 1][n - f1 - 1]: c = n - f1 - 1 r = binarySearchColumnInc(arr, n - f1 - 1, f1, n - f1 - 2, x) elif 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 or r == -1: print("-1") else: print("{0} {1}".format(r, c)) # Driver code if __name__ == "__main__": n = 4 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 Rituraj Jain |
C#
// C# implementation of the above approach using System;class GFG { 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 particular 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 // particular 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) { Console.Write("-1"); return; } // Edge case if n is odd if (n % 2 == 1 && f1 == (n + 1) / 2 - 1) { Console.WriteLine(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) Console.Write("-1"); else Console.Write(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 Arnab Kundu |
PHP
<?php// PHP implementation of the above approach$n = 4;// Function to return the ring, the number x// belongs to.function findRing($arr, $x){ global $n; // Returns -1 if number x is smaller than // least element of arr if ($arr[0][0] > $x) return -1; // l and r represent the diagonal // elements to search in $l = 0; $r = (int)(($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) { $mid = (int)(($l + $r) / 2); if ($arr[$mid][$mid] <= $x) if ($mid == (int)(($n + 1) / 2 - 1) || $arr[$mid + 1][$mid + 1] > $x) return $mid; else $l = $mid + 1; else $r = $mid - 1; } return $r;}// Function to perform binary search // on an array sorted in increasing order// l and r represent the left and right // index of the row to be searchedfunction binarySearchRowInc($arr, $row, $l, $r, $x){ while ($l <= $r) { $mid = (int)(($l + $r) / 2); if ($arr[$row][$mid] == $x) return $mid; if ($arr[$row][$mid] < $x) $l = $mid + 1; else $r = $mid - 1; } return -1;}// Function to perform binary search on // a particular column of the 2D array// t and b represent top and // bottom rowsfunction binarySearchColumnInc($arr, $col, $t, $b, $x){ while ($t <= $b) { $mid = (int)(($t + b) / 2); if ($arr[$mid][$col] == $x) return $mid; if ($arr[$mid][$col] < $x) $t = $mid + 1; else $b = $mid - 1; } return -1;}// Function to perform binary search on // an array sorted in decreasing orderfunction binarySearchRowDec($arr, $row, $l, $r, $x){ while ($l <= $r) { $mid = (int)(($l + $r) / 2); if ($arr[$row][$mid] == $x) return $mid; if ($arr[$row][$mid] < $x) $r = $mid - 1; else $l = $mid + 1; } return -1;}// Function to perform binary search on a// particular column of the 2D arrayfunction binarySearchColumnDec($arr, $col, $t, $b, $x){ while ($t <= $b) { $mid = (int)(($t + $b) / 2); if ($arr[$mid][$col] == $x) return $mid; if ($arr[$mid][$col] < $x) $b = $mid - 1; else $t = $mid + 1; } return -1;}// Function to find the position of the number xfunction spiralBinary($arr, $x){ global $n; // Finding the ring $f1 = findRing($arr, $x); // To store row and column $r = -1; $c = -1; if ($f1 == -1) { echo "-1"; return; } // Edge case if n is odd if ($n % 2 == 1 && $f1 == (int)(($n + 1) / 2 - 1)) { echo $f1 . " " . $f1 . "\n"; return; } // Check which of the 4 sides, the number x // lies in if ($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) echo "-1"; else echo $r . " " . $c; return;}// Driver code$arr = array(array( 1, 2, 3, 4 ), array( 12, 13, 14, 5 ), array( 11, 16, 15, 6 ), array( 10, 9, 8, 7 ));spiralBinary($arr, 7);// This code is contributed by mits?> |
Javascript
<script>// Javascript implementation of the above approachvar n = 4;// Function to return the ring, the number x// belongs to.function findRing(arr, 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 var l = 0, r = parseInt((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) { var mid = parseInt((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 searchedfunction binarySearchRowInc(arr, row, l, r, x){ while (l <= r) { var mid = parseInt((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 particular column of the 2D array// t and b represent top and // bottom rowsfunction binarySearchColumnInc(arr, col, t, b, x){ while (t <= b) { var mid = parseInt((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 orderfunction binarySearchRowDec(arr, row, l, r, x){ while (l <= r) { var mid = parseInt((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// particular column of the 2D arrayfunction binarySearchColumnDec(arr, col, t, b, x){ while (t <= b) { var mid = parseInt((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 xfunction spiralBinary(arr, x){ // Finding the ring var f1 = findRing(arr, x); // To store row and column var r, c; if (f1 == -1) { document.write( "-1"); return; } // Edge case if n is odd if (n % 2 == 1 && f1 == (n + 1) / 2 - 1) { document.write( f1 + " " + f1 + "<br>"); 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) document.write( "-1"); else document.write( r + " " + c); return;}// Driver codevar arr = [ [ 1, 2, 3, 4 ], [ 12, 13, 14, 5 ], [ 11, 16, 15, 6 ], [ 10, 9, 8, 7 ] ];spiralBinary(arr, 7);</script> |
Output:
3 3
Time Complexity: O(logN)
Auxiliary Space: O(logN)
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