Print K’th element in spiral form of matrix

Given a 2D Matrix of order n X m, print K’th element in the spiral form of the matrix. See the following examples.
Examples:

Input: mat[][] = 
          {{1, 2, 3, 4}
           {5, 6, 7, 8}
           {9, 10, 11, 12}
           {13, 14, 15, 16}}
       k = 6
Output: 12
Explanation: The elements in spiral order is 
1, 2, 3, 4, 8, 12, 16, 15...
so the 6th element is 12

Input: mat[][] =
       {{1, 2, 3, 4, 5, 6}
        {7, 8, 9, 10, 11, 12}
        {13, 14, 15, 16, 17, 18}}
       k = 17
Output: 10
Explanation: The elements in spiral order is 
1, 2, 3, 4, 5, 6, 12, 18, 17,
16, 15, 14, 13, 7, 8, 9, 10, 11 
so the 17 th element is 10.  

Simple Approah: One simple solution is to start traversing matrix in spiral form Print Spiral Matrix and start a counter i.e; count = 0. Whenever count gets equal to K, print that element.

• Algorithm:
  1. Keep a variable count = 0 to store the count.
  2. Traverse the matrix in sprial from start to end.
  3. Increase the count by 1 for every iteration.
  4. If the count is equal to the given value of k print the element and break.
• Implementation
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  #include <bits/stdc++.h> using namespace std; #define R 3 #define C 6    void spiralPrint(int m, int n, int a[R][C], int c) {     int i, k = 0, l = 0;     int count = 0;        /* k - starting row index          m - ending row index          l - starting column index          n - ending column index          i - iterator      */        while (k < m && l < n) {         /* check the first row from              the remaining rows */         for (i = l; i < n; ++i) {             count++;                if (count == c)                 cout << a[k][i] << " ";         }         k++;            /* check the last column          from the remaining columns */         for (i = k; i < m; ++i) {             count++;                if (count == c)                 cout << a[i][n - 1] << " ";         }         n--;            /* check the last row from                  the remaining rows */         if (k < m) {             for (i = n - 1; i >= l; --i) {                 count++;                    if (count == c)                     cout << a[m - 1][i] << " ";             }             m--;         }            /* check the first column from                  the remaining columns */         if (l < n) {             for (i = m - 1; i >= k; --i) {                 count++;                    if (count == c)                     cout << a[i][l] << " ";             }             l++;         }     } }    /* Driver program to test above functions */ int main() {     int a[R][C] = { { 1, 2, 3, 4, 5, 6 },                     { 7, 8, 9, 10, 11, 12 },                     { 13, 14, 15, 16, 17, 18 } },         k = 17;        spiralPrint(R, C, a, k);     return 0; }

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  Output:

  10
  

• Complexity Analysis:
  • Time Complexity : O(R*C), A single traversal of matrix is needed so the Time Complexity is O(R*C).
  • Space Complexity : O(1), constant space is required.

Efficient Approach: While traversing the array in spiral order, a loop is used to traverse the sides. So if it can be found out that the kth element is in the given side then the kth element can be found out in constant time. This can be done recursively as well as iteratively.

• Algorithm :
  1. Traverse the matrix in form of spiral or cycles.
  2. So a cycle can be divided into 4 parts, so if the cycle is of size m X n.
  3. Element is in first row, i.e k <= m
  4. Element is in last column, i.e k <= (m+n-1)
  5. Element is in last row, i.e. k <= (m+n-1+m-1)
  6. Element is in first column, i.e k <= (m+n-1+m-1+n-2)
  7. If any of the above conditions meet then the kth element can be found is constant time.
  8. Else remove the cycle from the array and recursively call the function.
• Implementation:
  

  C++

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  // C++ program for Kth element in spiral // form of matrix #include <bits/stdc++.h> #define MAX 100 using namespace std;    /* function for Kth element */ int findK(int A[MAX][MAX], int n, int m, int k) {     if (n < 1 || m < 1)         return -1;        /*....If element is in outermost ring ....*/     /* Element is in first row */     if (k <= m)         return A[0][k - 1];        /* Element is in last column */     if (k <= (m + n - 1))         return A[(k - m)][m - 1];        /* Element is in last row */     if (k <= (m + n - 1 + m - 1))         return A[n - 1][m - 1 - (k - (m + n - 1))];        /* Element is in first column */     if (k <= (m + n - 1 + m - 1 + n - 2))         return A[n - 1 - (k - (m + n - 1 + m - 1))][0];        /*....If element is NOT in outermost ring ....*/     /* Recursion for sub-matrix. &A[1][1] is     address to next inside sub matrix.*/     return findK((int(*)[MAX])(&(A[1][1])), n - 2,                  m - 2, k - (2 * n + 2 * m - 4)); }    /* Driver code */ int main() {     int a[MAX][MAX] = { { 1, 2, 3, 4, 5, 6 },                         { 7, 8, 9, 10, 11, 12 },                         { 13, 14, 15, 16, 17, 18 } };     int k = 17;     cout << findK(a, 3, 6, k) << endl;     return 0; }

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  Java

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  // Java program for Kth element in spiral // form of matrix class GFG {        static int MAX = 100;        /* function for Kth element */     static int findK(int A[][], int i, int j,                      int n, int m, int k)     {         if (n < 1 || m < 1)             return -1;            /*.....If element is in outermost ring ....*/         /* Element is in first row */         if (k <= m)             return A[i + 0][j + k - 1];            /* Element is in last column */         if (k <= (m + n - 1))             return A[i + (k - m)][j + m - 1];            /* Element is in last row */         if (k <= (m + n - 1 + m - 1))             return A[i + n - 1][j + m - 1 - (k - (m + n - 1))];            /* Element is in first column */         if (k <= (m + n - 1 + m - 1 + n - 2))             return A[i + n - 1 - (k - (m + n - 1 + m - 1))][j + 0];            /*.....If element is NOT in outermost ring ....*/         /* Recursion for sub-matrix. &A[1][1] is     address to next inside sub matrix.*/         return findK(A, i + 1, j + 1, n - 2,                      m - 2, k - (2 * n + 2 * m - 4));     }        /* Driver code */     public static void main(String args[])     {         int a[][] = { { 1, 2, 3, 4, 5, 6 },                       { 7, 8, 9, 10, 11, 12 },                       { 13, 14, 15, 16, 17, 18 } };         int k = 17;         System.out.println(findK(a, 0, 0, 3, 6, k));     } }    // This code is contributed by Arnab Kundu

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  Output:

   10
  

• Complexity Analysis:
  • Time Complexity: O(c), where c is number of outer circular rings with respect to k’th element.
  • Space Complexity: O(1).
    As constant space is required.

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