Given a matrix arr[][] of dimensions N * M and an integer K, the task is to print all elements of the matrix starting from the top-left element up to K diagonally in spiral form.
Examples:
Input : N=5, M=6, K=15, arr[][]={{1, 2, 3, 4, 5, 6},
{7, 8, 9, 10, 11, 12},
{13, 14, 15, 16, 17, 18},
{19, 20, 21, 22, 23, 24},
{25, 26, 27, 28, 29, 30}}
Output: 1, 2, 7, 13, 8, 3, 4, 9, 14, 19, 25, 20, 15
Explanation:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1st diagonal printed: {1}
2nd diagonal printed: {2, 7}
3rd diagonal printed: {13, 8, 3}
……
5th diagonal printed {25, 20, 15}.
Since 15 is encountered, no further matrix element is printed.Input: N = 4, M = 3, K = 69, arr[][]={{4, 87, 24},
{17, 1, 18},
{25, 69, 97},
{19, 27, 85}}
Output: 4, 87, 17, 25, 1, 24, 18, 69
Approach: Follow the steps below to solve the problem:
- The total number of diagonals in the matrix is N + M – 1.
- Traverse the diagonal one by one in spiral manner.
- For every element traversed, check if it is equal to K or not. If found to be true, print that element and terminate.
- Otherwise, print the element and evaluate the next indices to be traversed. If i and j are the current indexes:
- While moving diagonally up i will be decremented and j will be incremented.
- While moving diagonally down i will be incremented and j will be decremented.
- If the next index is not a valid index, then move to the next diagonal.
- Otherwise, update the current position to the next.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to check if the // indices are valid or not bool isValid(int i, int j, int N, int M) { return (i >= 0 && i < N && j >= 0 && j < M); } // Function to evaluate the next // index while moving diagonally up pair<int, int> up(int i, int j, int N, int M) { if (isValid(i - 1, j + 1, N, M)) return { i - 1, j + 1 }; else return { -1, -1 }; } // Function to evaluate the next // index while moving diagonally down pair<int, int> down(int i, int j, int N, int M) { if (isValid(i + 1, j - 1, N, M)) return { i + 1, j - 1 }; else return { -1, -1 }; } // Function to print matrix elements // diagonally in Spiral Form void SpiralDiagonal(int N, int M, int K, vector<vector<int> > a) { int i = 0, j = 0; // Total Number of Diagonals // = N + M - 1 for (int diagonal = 0; diagonal < N + M - 1; diagonal++) { while (1) { // Stop when K is // encountered if (a[i][j] == K) { cout << K; return; } // Print the integer cout << a[i][j] << ", "; // Store the next index pair<int, int> next; if (diagonal & 1) { next = down(i, j, N, M); } else { next = up(i, j, N, M); } // If current index is invalid if (next.first == next.second && next.first == -1) { // Move to the next diagonal if (diagonal & 1) { (i + 1 < N) ? ++i : ++j; } else { (j + 1 < M) ? ++j : ++i; } break; } // Otherwise move to the // next valid index else { i = next.first; j = next.second; } } } } // Driver Code int main() { int N = 5, M = 6, K = 15; vector<vector<int> > arr = { { 1, 2, 3, 4, 5, 6 }, { 7, 8, 9, 10, 11, 12 }, { 13, 14, 15, 16, 17, 18 }, { 19, 20, 21, 22, 23, 24 }, { 25, 26, 27, 28, 29, 30 } }; // Function Call SpiralDiagonal(N, M, K, arr); return 0; } |
Java
// Java program for the // above approachimport java.util.*;import java.lang.*; class GFG{static class pair{ int first, second; pair(int f, int s) { this.first = f; this.second = s; }}// Function to check if the// indices are valid or notstatic boolean isValid(int i, int j, int N, int M){ return (i >= 0 && i < N && j >= 0 && j < M);} // Function to evaluate the next// index while moving diagonally upstatic int[] up(int i, int j, int N, int M){ if (isValid(i - 1, j + 1, N, M)) return new int[]{ i - 1, j + 1 }; else return new int[]{ -1, -1 };} // Function to evaluate the next// index while moving diagonally downstatic int[] down(int i, int j, int N, int M){ if (isValid(i + 1, j - 1, N, M)) return new int[]{ i + 1, j - 1 }; else return new int[]{ -1, -1 };} // Function to print matrix elements// diagonally in Spiral Formstatic void SpiralDiagonal(int N, int M, int K, int[][] a){ int i = 0, j = 0; // Total Number of Diagonals // = N + M - 1 for(int diagonal = 0; diagonal < N + M - 1; diagonal++) { while (true) { // Stop when K is // encountered if (a[i][j] == K) { System.out.print(K); return; } // Print the integer System.out.print(a[i][j] + ", "); // Store the next index int[] next; if ((diagonal & 1) == 1) { next = down(i, j, N, M); } else { next = up(i, j, N, M); } // If current index is invalid if (next[0] == next[1] && next[1] == -1) { // Move to the next diagonal if ((diagonal & 1) == 1) { if (i + 1 < N) ++i; else ++j; } else { if (j + 1 < M) ++j; else ++i; } break; } // Otherwise move to the // next valid index else { i = next[0]; j = next[1]; } } }} // Driver codepublic static void main (String[] args){ int N = 5, M = 6, K = 15; int[][] arr = { { 1, 2, 3, 4, 5, 6 }, { 7, 8, 9, 10, 11, 12 }, { 13, 14, 15, 16, 17, 18 }, { 19, 20, 21, 22, 23, 24 }, { 25, 26, 27, 28, 29, 30 } }; // Function Call SpiralDiagonal(N, M, K, arr);}}// This code is contributed by offbeat |
Python3
# Python3 program for the # above approach# Function to check if the# indices are valid or notdef isValid(i, j, N, M): return (i >= 0 and i < N and j >= 0 and j < M)# Function to evaluate the next# index while moving diagonally updef up(i, j, N, M): if(isValid(i - 1, j + 1, N, M)): return [i - 1, j + 1 ] else: return [-1, -1]# Function to evaluate the next# index while moving diagonally downdef down(i, j, N, M): if(isValid(i + 1, j - 1, N, M)): return [i + 1, j - 1 ] else: return [-1, -1]# Function to print matrix elements# diagonally in Spiral Formdef SpiralDiagonal(N, M, K, a): i = 0 j = 0 # Total Number of Diagonals # = N + M - 1 for diagonal in range(N + M - 1): while(True): # Stop when K is # encountered if(a[i][j] == K): print(K, end = "") return # Print the integer print(a[i][j], ", ", end="", sep="") # Store the next index next = [] if((diagonal & 1) == 1): next = down(i, j, N, M) else: next = up(i, j, N, M) # If current index is invalid if(next[0] == next[1] and next[1] == -1): # Move to the next diagonal if((diagonal & 1) == 1): if(i + 1 < N): i += 1 else: j += 1 else: if(j + 1 < M): j += 1 else: i += 1 break # Otherwise move to the # next valid index else: i = next[0] j = next[1]# Driver codeN = 5M = 6K = 15arr = [[1, 2, 3, 4, 5, 6 ], [ 7, 8, 9, 10, 11, 12], [13, 14, 15, 16, 17, 18 ], [19, 20, 21, 22, 23, 24], [25, 26, 27, 28, 29, 30]]# Function CallSpiralDiagonal(N, M, K, arr);# This code is contributed by avanitrachhadiya2155 |
C#
// C# program for the// above approachusing System;class GFG{// Function to check if the// indices are valid or notstatic bool isValid(int i, int j, int N, int M){ return (i >= 0 && i < N && j >= 0 && j < M);}// Function to evaluate the next// index while moving diagonally upstatic int[] up(int i, int j, int N, int M){ if (isValid(i - 1, j + 1, N, M)) return new int[]{ i - 1, j + 1 }; else return new int[]{ -1, -1 };}// Function to evaluate the next// index while moving diagonally downstatic int[] down(int i, int j, int N, int M){ if (isValid(i + 1, j - 1, N, M)) return new int[]{ i + 1, j - 1 }; else return new int[]{ -1, -1 };}// Function to print matrix elements// diagonally in Spiral Formstatic void SpiralDiagonal(int N, int M, int K, int[, ] a){ int i = 0, j = 0; // Total Number of Diagonals // = N + M - 1 for(int diagonal = 0; diagonal < N + M - 1; diagonal++) { while (true) { // Stop when K is // encountered if (a[i, j] == K) { Console.Write(K); return; } // Print the integer Console.Write(a[i, j] + ", "); // Store the next index int[] next; if ((diagonal & 1) == 1) { next = down(i, j, N, M); } else { next = up(i, j, N, M); } // If current index is invalid if (next[0] == next[1] && next[1] == -1) { // Move to the next diagonal if ((diagonal & 1) == 1) { if (i + 1 < N) ++i; else ++j; } else { if (j + 1 < M) ++j; else ++i; } break; } // Otherwise move to the // next valid index else { i = next[0]; j = next[1]; } } }}// Driver codepublic static void Main(string[] args){ int N = 5, M = 6, K = 15; int[, ] arr = { { 1, 2, 3, 4, 5, 6 }, { 7, 8, 9, 10, 11, 12 }, { 13, 14, 15, 16, 17, 18 }, { 19, 20, 21, 22, 23, 24 }, { 25, 26, 27, 28, 29, 30 } }; // Function Call SpiralDiagonal(N, M, K, arr);}}// This code is contributed by grand_master |
Javascript
<script>// JavaScript implementation for the above approach // Function to check if the // indices are valid or not function isValid(i, j, N, M) { return (i >= 0 && i < N && j >= 0 && j < M); } // Function to evaluate the next // index while moving diagonally up function up( i, j, N, M) { if (isValid(i - 1, j + 1, N, M)) return [ i - 1, j + 1 ]; else return [ -1, -1 ]; } // Function to evaluate the next // index while moving diagonally down function down( i, j, N, M) { if (isValid(i + 1, j - 1, N, M)) return [ i + 1, j - 1 ]; else return [ -1, -1 ]; } // Function to print matrix elements // diagonally in Spiral Form function SpiralDiagonal(N,M,K,a) { var i = 0, j = 0; // Total Number of Diagonals // = N + M - 1 for (var diagonal = 0; diagonal < N + M - 1; diagonal++) { while (1) { // Stop when K is // encountered if (a[i][j] == K) { document.write(K); return; } // Print the integer document.write(a[i][j], ", "); // Store the next index var next = new Array(2); if (diagonal & 1) { next = down(i, j, N, M); } else { next = up(i, j, N, M); } // If current index is invalid if (next[0] == next[1] && next[0] == -1) { // Move to the next diagonal if (diagonal & 1) { (i + 1 < N) ? ++i : ++j; } else { (j + 1 < M) ? ++j : ++i; } break; } // Otherwise move to the // next valid index else { i = next[0]; j = next[1]; } } } } // Driver Code var N = 5, M = 6, K = 15; var arr = [[1, 2, 3, 4, 5, 6 ], [ 7, 8, 9, 10, 11, 12], [13, 14, 15, 16, 17, 18 ], [19, 20, 21, 22, 23, 24], [25, 26, 27, 28, 29, 30]]; // Function Call SpiralDiagonal(N, M, K, arr); // This code is contributed by Shubham Singh</script> |
Output:
1, 2, 7, 13, 8, 3, 4, 9, 14, 19, 25, 20, 15
Time Complexity: O(N*M)
Auxiliary Space: O(1)
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