Given a number N, the task is to write a C program to find the square root of the given number N.
Examples:
Input: N = 12
Output: 3.464102Input: N = 16
Output: 4
Method 1: Using inbuilt sqrt() function: The sqrt() function returns the sqrt of any number N.
Below is the implementation of the above approach:
C
// C program for the above approach#include <math.h>#include <stdio.h>// Function to find the square-root of Ndouble findSQRT(double N) { return sqrt(N); }// Driver Codeint main(){ // Given number int N = 12; // Function call printf("%f ", findSQRT(N)); return 0;} |
3.464102
Time complexity: O(logN), as the inbuilt sqrt() function take log(n)
Auxiliary space: O(1)
Method 2: Using Binary Search: This approach is used to find the square root of the given number N with precision upto 5 decimal places.
- The square root of number N lies in range 0 ? squareRoot ? N. Initialize start = 0 and end = number.
- Compare the square of the mid integer with the given number. If it is equal to the number, then we found our integral part, else look for the same in the left or right side of mid depending upon the condition.
- After finding an integral part, we will find the fractional part.
- Initialize the increment variable by 0.1 and iteratively calculate the fractional part upto 5 decimal places.
- For each iteration, change increment to 1/10th of its previous value.
- Finally, return the answer computed.
Below is the implementation of the above approach:
C
// C program for the above approach#include <stdio.h>#include <stdlib.h>// Function to find the square-root of Nfloat findSQRT(int number){ int start = 0, end = number; int mid; // To store the answer float ans; // To find integral part of square // root of number while (start <= end) { // Find mid mid = (start + end) / 2; // If number is perfect square // then break if (mid * mid == number) { ans = mid; break; } // Increment start if integral // part lies on right side // of the mid if (mid * mid < number) { //first start value should be added to answer ans=start; //then start should be changed start = mid + 1; } // Decrement end if integral part // lies on the left side of the mid else { end = mid - 1; } } // To find the fractional part // of square root upto 5 decimal float increment = 0.1; for (int i = 0; i < 5; i++) { while (ans * ans <= number) { ans += increment; } // Loop terminates, // when ans * ans > number ans = ans - increment; increment = increment / 10; } return ans;}// Driver Codeint main(){ // Given number int N = 12; // Function call printf("%f ", findSQRT(N)); return 0;} |
3.464099
Method 3: Using log2(): The square-root of a number N can be calculated using log2() as:
Let d be our answer for input number N, then
Apply log2() both sides
Therefore,
d = pow(2, 0.5*log2(n))
Below is the implementation of the above approach:
C
// C program for the above approach#include <math.h>#include <stdio.h>// Function to find the square-root of Ndouble findSQRT(double N) { return pow(2, 0.5 * log2(N)); }// Driver Codeint main(){ // Given number int N = 12; // Function call printf("%f ", findSQRT(N)); return 0;} |
3.464102
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