Given an infinite matrix filled with the natural numbers as shown below:
1 2 4 7 . . . 3 5 8 . . . . 6 9 . . . . . 10 . . . . . . . . . . . .
Also, given an integer N and the task is to find the row and the column of the integer N in the given matrix.
Examples:
Input: N = 5 Output: 2 2 5 is present in the 2nd row and the 2nd column. Input: N = 3 Output: 2 1
Approach: On observing the problem carefully, the row number can be obtained by subtracting first x natural numbers from N such that they satisfy the condition N – (x * (x + 1)) / 2 ? 1 and the resultant value will be the required row number.
To get the corresponding column, add first x natural numbers to 1 such that they satisfy the condition 1 + (y * (y + 1)) / 2 ? A. Subtract this resultant value from N to get the gap between the base and the given value and again subtract the gap from y + 1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the row and// the column of the given integerpair<int, int> solve(int n){ int low = 1, high = 1e4, x = n, p = 0; // Binary search for the row number while (low <= high) { int mid = (low + high) / 2; int sum = (mid * (mid + 1)) / 2; // Condition to get the maximum // x that satisfies the criteria if (x - sum >= 1) { p = mid; low = mid + 1; } else { high = mid - 1; } } int start = 1, end = 1e4, y = 1, q = 0; // Binary search for the column number while (start <= end) { int mid = (start + end) / 2; int sum = (mid * (mid + 1)) / 2; // Condition to get the maximum // y that satisfies the criteria if (y + sum <= n) { q = mid; start = mid + 1; } else { end = mid - 1; } } // Get the row and the column number x = x - (p * (p + 1)) / 2; y = y + (q * (q + 1)) / 2; int r = x; int c = q + 1 - n + y; // Return the pair pair<int, int> ans = { r, c }; return ans;}// Driver codeint main(){ int n = 5; pair<int, int> p = solve(n); cout << p.first << " " << p.second; return 0;} |
Java
// Java implementation of the approach class GFG { // Function to return the row and // the column of the given integer static int[] solve(int n) { int low = 1, high = (int)1e4, x = n, p = 0; // Binary search for the row number while (low <= high) { int mid = (low + high) / 2; int sum = (mid * (mid + 1)) / 2; // Condition to get the maximum // x that satisfies the criteria if (x - sum >= 1) { p = mid; low = mid + 1; } else { high = mid - 1; } } int start = 1, end = (int)1e4, y = 1, q = 0; // Binary search for the column number while (start <= end) { int mid = (start + end) / 2; int sum = (mid * (mid + 1)) / 2; // Condition to get the maximum // y that satisfies the criteria if (y + sum <= n) { q = mid; start = mid + 1; } else { end = mid - 1; } } // Get the row and the column number x = x - (p * (p + 1)) / 2; y = y + (q * (q + 1)) / 2; int r = x; int c = q + 1 - n + y; // Return the pair int ans[] = { r, c }; return ans; } // Driver code public static void main (String[] args) { int n = 5; int []p = solve(n); System.out.println(p[0] + " " + p[1]); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach# Function to return the row and# the column of the given integerdef solve(n): low = 1 high = 10**4 x, p = n, 0 # Binary search for the row number while (low <= high): mid = (low + high) // 2 sum = (mid * (mid + 1)) // 2 # Condition to get the maximum # x that satisfies the criteria if (x - sum >= 1): p = mid low = mid + 1 else : high = mid - 1 start, end, y, q = 1, 10**4, 1, 0 # Binary search for the column number while (start <= end): mid = (start + end) // 2 sum = (mid * (mid + 1)) // 2 # Condition to get the maximum # y that satisfies the criteria if (y + sum <= n): q = mid start = mid + 1 else: end = mid - 1 # Get the row and the column number x = x - (p * (p + 1)) // 2 y = y + (q * (q + 1)) // 2 r = x c = q + 1 - n + y # Return the pair return r, c# Driver coden = 5r, c = solve(n)print(r, c)# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System;class GFG { // Function to return the row and // the column of the given integer static int[] solve(int n) { int low = 1, high = (int)1e4, x = n, p = 0; // Binary search for the row number while (low <= high) { int mid = (low + high) / 2; int sum = (mid * (mid + 1)) / 2; // Condition to get the maximum // x that satisfies the criteria if (x - sum >= 1) { p = mid; low = mid + 1; } else { high = mid - 1; } } int start = 1, end = (int)1e4, y = 1, q = 0; // Binary search for the column number while (start <= end) { int mid = (start + end) / 2; int sum = (mid * (mid + 1)) / 2; // Condition to get the maximum // y that satisfies the criteria if (y + sum <= n) { q = mid; start = mid + 1; } else { end = mid - 1; } } // Get the row and the column number x = x - (p * (p + 1)) / 2; y = y + (q * (q + 1)) / 2; int r = x; int c = q + 1 - n + y; // Return the pair int []ans = {r, c}; return ans; } // Driver code public static void main (String[] args) { int n = 5; int []p = solve(n); Console.WriteLine(p[0] + " " + p[1]); } }// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript implementation of the approach // Function to return the row and // the column of the given integer function solve(n){ let low = 1, high = 1e4, x = n, p = 0; // Binary search for the row number while (low <= high) { let mid = Math.floor((low + high) / 2); let sum = Math.floor((mid * (mid + 1)) / 2); // Condition to get the maximum // x that satisfies the criteria if (x - sum >= 1) { p = mid; low = mid + 1; } else { high = mid - 1; } } let start = 1, end = 1e4, y = 1, q = 0; // Binary search for the column number while (start <= end) { let mid = Math.floor((start + end) / 2); let sum = Math.floor((mid * (mid + 1)) / 2); // Condition to get the maximum // y that satisfies the criteria if (y + sum <= n) { q = mid; start = mid + 1; } else { end = mid - 1; } } // Get the row and the column number x = x - Math.floor((p * (p + 1)) / 2); y = y + Math.floor((q * (q + 1)) / 2); let r = x; let c = q + 1 - n + y; // Return the pair let ans = [ r, c ]; return ans; }// Driver code let n = 5; let p = solve(n); document.write(p[0] + " " + p[1]); // This code is contributed by patel2127</script> |
Output:
2 2
Time Complexity: O(log(N))
Auxiliary Space: O(1)
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