Find N in the given matrix that follows a pattern

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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 2^nd row 
and the 2^nd 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 integer
pair<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 code
int 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 integer
def 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 code
n = 5
 
r, 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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