Given an N * M grid. It contains exactly three ‘*’ and every other cell is a ‘.’ where ‘*’ denotes a vertex of a rectangle. The task is to find the coordinates of the fourth (missing) vertex (1-based indexing).
Examples:
Input: grid[][] = {“*.*”, “*..”, “…”}
Output: 2 3
(1, 1), (1, 3) and (2, 1) are the given coordinates of the rectangle where (2, 3) is the missing coordinate.
Input: grid[][] = {“*.*”, “..*”}
Output: 2 1
Approach: Find the coordinates of the 3 vertices by iterating through the given grid. Since a rectangle consists of 2 X-coordinates and 2 Y-coordinates and 4 vertices, every X-coordinate and Y-coordinate should occur exactly twice. We can count how many times each X and Y coordinate occurs in the 3 given vertices and the 4th one will have coordinates that occur only once.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that return the coordinates// of the fourth vertex of the rectanglepair<int, int> findFourthVertex(int n, int m, string s[]){ unordered_map<int, int> row, col; for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) // Save the coordinates of the given // vertices of the rectangle if (s[i][j] == '*') { row[i]++; col[j]++; } int x, y; for (auto tm : row) if (tm.second == 1) x = tm.first; for (auto tm : col) if (tm.second == 1) y = tm.first; // 1-based indexing return make_pair(x + 1, y + 1);}// Driver codeint main(){ string s[] = { "*.*", "*..", "..." }; int n = sizeof(s) / sizeof(s[0]); int m = s[0].length(); auto rs = findFourthVertex(n, m, s); cout << rs.first << " " << rs.second;} |
Java
// Java implementation of the approachimport java.util.HashMap;import java.util.Map;class GfG{ // Function that return the coordinates // of the fourth vertex of the rectangle static Pair<Integer, Integer> findFourthVertex(int n, int m, String s[]) { HashMap<Integer, Integer> row = new HashMap<>(); HashMap<Integer, Integer> col = new HashMap<>(); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { // Save the coordinates of the given // vertices of the rectangle if (s[i].charAt(j) == '*') { if (row.containsKey(i)) { row.put(i, row.get(i) + 1); } else { row.put(i, 1); } if (col.containsKey(j)) { col.put(j, col.get(j) + 1); } else { col.put(j, 1); } } } } int x = 0, y = 0; for (Map.Entry<Integer, Integer> entry : row.entrySet()) { if (entry.getValue() == 1) x = entry.getKey(); } for (Map.Entry<Integer, Integer> entry : col.entrySet()) { if (entry.getValue() == 1) y = entry.getKey(); } // 1-based indexing Pair<Integer, Integer> ans = new Pair<>(x + 1, y + 1); return ans; } // Driver Code public static void main(String []args) { String s[] = { "*.*", "*..", "..." }; int n = s.length; int m = s[0].length(); Pair<Integer, Integer> rs = findFourthVertex(n, m, s); System.out.println(rs.first + " " + rs.second); }}class Pair<A, B> { A first; B second; public Pair(A first, B second) { this.first = first; this.second = second; }} // This code is contributed by Rituraj Jain |
Python3
# Python3 implementation of the approach # Function that return the coordinates # of the fourth vertex of the rectangle def findFourthVertex(n, m, s) : row = dict.fromkeys(range(n), 0) col = dict.fromkeys(range(m), 0) for i in range(n) : for j in range(m) : # Save the coordinates of the given # vertices of the rectangle if (s[i][j] == '*') : row[i] += 1; col[j] += 1; for keys,values in row.items() : if (values == 1) : x = keys; for keys,values in col.items() : if (values == 1) : y = keys; # 1-based indexing return (x + 1, y + 1) ;# Driver code if __name__ == "__main__" : s = [ "*.*", "*..", "..." ] n = len(s); m = len(s[0]); rs = findFourthVertex(n, m, s); print(rs[0], rs[1]) # This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GfG{ // Function that return the coordinates // of the fourth vertex of the rectangle static Pair<int, int> findFourthVertex(int n, int m, String []s) { Dictionary<int, int> row = new Dictionary<int, int>(); Dictionary<int, int> col = new Dictionary<int, int>(); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { // Save the coordinates of the given // vertices of the rectangle if (s[i][j] == '*') { if (row.ContainsKey(i)) { row[i] = row[i] + 1; } else { row.Add(i, 1); } if (col.ContainsKey(j)) { col[j] = col[j] + 1; } else { col.Add(j, 1); } } } } int x = 0, y = 0; foreach(KeyValuePair<int, int> entry in row) { if (entry.Value == 1) x = entry.Key; } foreach(KeyValuePair<int, int> entry in col) { if (entry.Value == 1) y = entry.Key; } // 1-based indexing Pair<int, int> ans = new Pair<int, int>(x + 1, y + 1); return ans; } // Driver Code public static void Main(String []args) { String []s = { "*.*", "*..", "..." }; int n = s.Length; int m = s[0].Length; Pair<int, int> rs = findFourthVertex(n, m, s); Console.WriteLine(rs.first + " " + rs.second); }}class Pair<A, B> { public A first; public B second; public Pair(A first, B second) { this.first = first; this.second = second; }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function that return the coordinates// of the fourth vertex of the rectanglefunction findFourthVertex(n, m, s){ var row = new Map(), col = new Map(); for (var i = 0; i < n; i++) for (var j = 0; j < m; j++) // Save the coordinates of the given // vertices of the rectangle if (s[i][j] == '*') { if(row.has(i)) row.set(i, row.get(i)+1) else row.set(i, 1) if(col.has(j)) col.set(j, col.get(j)+1) else col.set(j, 1) } var x, y; row.forEach((value, key) => { if (value == 1) x = key; }); col.forEach((value, key) => { if (value == 1) y = key; }); // 1-based indexing return [x + 1, y + 1];}// Driver codevar s = ["*.*", "*..", "..." ];var n = s.length;var m = s[0].length;var rs = findFourthVertex(n, m, s);document.write( rs[0] + " " + rs[1]);// This code is contributed by famously.</script> |
Output:
2 3
Time Complexity: O(N*M), as we are using a loop to traverse N*M times.
Auxiliary Space: O(N + M), as we are using extra space for map.
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