Prerequisite: Asymptotic Analysis, Worst, Average and Best Cases, Asymptotic Notations, Analysis of loops.
Problem 1: Find the complexity of the below recurrence:
{ 3T(n-1), if n>0,
T(n) = { 1, otherwise
Solution:
Let us solve using substitution.
T(n) = 3T(n-1)
= 3(3T(n-2))
= 32T(n-2)
= 33T(n-3)
…
…
= 3nT(n-n)
= 3nT(0)
= 3nThis clearly shows that the complexity of this function is O(3n).
Problem 2: Find the complexity of the recurrence:
{ 2T(n-1) – 1, if n>0,
T(n) = { 1, otherwise
Solution:
Let us try solving this function with substitution.
T(n) = 2T(n-1) – 1
= 2(2T(n-2)-1)-1
= 22(T(n-2)) – 2 – 1
= 22(2T(n-3)-1) – 2 – 1
= 23T(n-3) – 22 – 21 – 20
…..
…..
= 2nT(n-n) – 2n-1 – 2n-2 – 2n-3
….. 22 – 21 – 20= 2n – 2n-1 – 2n-2 – 2n-3
….. 22 – 21 – 20
= 2n – (2n-1)[Note: 2n-1 + 2n-2 + …… + 20 = 2n – 1]
T(n) = 1
Time Complexity is O(1). Note that while the recurrence relation looks exponential
he solution to the recurrence relation here gives a different result.
Problem 3: Find the complexity of the below program:
CPP
void function(int n){ if (n==1) return; for (int i=1; i<=n; i++) { for (int j=1; j<=n; j++) { cout << "*"; break; } cout << endl; }} |
Java
/*package whatever //do not write package name here */static void function(int n){ if (n==1) return; for (int i=1; i<=n; i++) { for (int j=1; j<=n; j++) { System.out.print("*"); break; } System.out.println(); }}// The code is contributed by Nidhi goel. |
Javascript
function funct(n){ if (n==1) return; for (let i=1; i<=n; i++) { for (let j=1; j<=n; j++) { console.log("*"); break; } console.log(); }}// The code is contributed by Nidhi goel. |
Python3
def funct(n): if (n==1): return for i in range(1, n+1): for j in range(1, n + 1): print("*", end = "") break print() # The code is contributed by Nidhi goel. |
C#
/*package whatever //do not write package name here */public static void function(int n){ if (n==1) return; for (int i=1; i<=n; i++) { for (int j=1; j<=n; j++) { Console.Write("*"); break; } Console.WriteLine(); }}// The code is contributed by Nidhi goel. |
Solution: Consider the comments in the following function.
CPP
function(int n){ if (n==1) return; for (int i=1; i<=n; i++) { // Inner loop executes only one // time due to break statement. for (int j=1; j<=n; j++) { printf("*"); break; } }} |
Time Complexity: O(n), Even though the inner loop is bounded by n, but due to the break statement, it is executing only once.
Problem 4: Find the complexity of the below program:
CPP
void function(int n){ int count = 0; for (int i=n/2; i<=n; i++) for (int j=1; j<=n; j = 2 * j) for (int k=1; k<=n; k = k * 2) count++;} |
Java
static void function(int n){ int count = 0; for (int i = n / 2; i <= n; i++) for (int j = 1; j <= n; j = 2 * j) for (int k = 1; k <= n; k = k * 2) count++;}// This code is contributed by rutvik_56. |
C#
static void function(int n){ int count = 0; for (int i = n / 2; i <= n; i++) for (int j = 1; j <= n; j = 2 * j) for (int k = 1; k <= n; k = k * 2) count++;}// This code is contributed by pratham76. |
Javascript
<script>function function1(n){ var count = 0; for (i = n / 2; i <= n; i++) for (j = 1; j <= n; j = 2 * j) for (k = 1; k <= n; k = k * 2) count++;}// This code is contributed by umadevi9616 </script> |
Solution: Consider the comments in the following function.
CPP
void function(int n){ int count = 0; for (int i=n/2; i<=n; i++) // Executes O(Log n) times for (int j=1; j<=n; j = 2 * j) // Executes O(Log n) times for (int k=1; k<=n; k = k * 2) count++;} |
Time Complexity: O(n log2n).
Problem 5: Find the complexity of the below program:
CPP
void function(int n){ int count = 0; for (int i=n/2; i<=n; i++) for (int j=1; j+n/2<=n; j = j++) for (int k=1; k<=n; k = k * 2) count++;} |
Java
static void function(int n){ int count = 0; for (int i=n/2; i<=n; i++) for (int j=1; j+n/2<=n; j = j++) for (int k=1; k<=n; k = k * 2) count++;}// This code is contributed by Pushpesh Raj. |
Solution: Consider the comments in the following function.
CPP
void function(int n){ int count = 0; // outer loop executes n/2 times for (int i=n/2; i<=n; i++) // middle loop executes n/2 times for (int j=1; j+n/2<=n; j = j++) // inner loop executes logn times for (int k=1; k<=n; k = k * 2) count++;}// The code is contributed by Nidhi goel. |
Java
static void function(int n){ int count = 0; // outer loop executes n/2 times for (int i=n/2; i<=n; i++) // middle loop executes n/2 times for (int j=1; j+n/2<=n; j = j++) // inner loop executes logn times for (int k=1; k<=n; k = k * 2) count++;}// This code is contributed by Aman Kumar |
Javascript
function function(n){ let count = 0; // outer loop executes n/2 times for (let i= Math.floor(n/2); i<=n; i++) // middle loop executes n/2 times for (let j=1; j+n/2<=n; j = j++) // inner loop executes logn times for (let k=1; k<=n; k = k * 2) count++;}// The code is contributed by Nidhi goel. |
Python3
def function(n): count = 0 # outer loop executes n/2 times for i in range(n//2, n+1): # middle loop executes n/2 times for j in range((1, n//2 + 1): # inner loop executes logn times for k in range(1, n+1, 2): count++# The code is contributed by Nidhi goel. |
C#
using System;public static void function(int n){ int count = 0; // outer loop executes n/2 times for (int i=n/2; i<=n; i++) // middle loop executes n/2 times for (int j=1; j+n/2<=n; j = j++) // inner loop executes logn times for (int k=1; k<=n; k = k * 2) count++;}// The code is contributed by Nidhi goel. |
Time Complexity: O(n2logn).
Problem 6: Find the complexity of the below program:
CPP
void function(int n){ int i = 1, s =1; while (s <= n) { i++; s += i; printf("*"); }} |
Solution: We can define the terms ‘s’ according to relation si = si-1 + i. The value of ‘i’ increases by one for each iteration. The value contained in ‘s’ at the ith iteration is the sum of the first ‘i’ positive integers. If k is total number of iterations taken by the program, then while loop terminates if: 1 + 2 + 3 ….+ k = [k(k+1)/2] > n So k = O(√n).
Time Complexity: O(√n).
Problem 7: Find a tight upper bound on the complexity of the below program:
CPP
void function(int n){ int count = 0; for (int i=0; i<n; i++) for (int j=i; j< i*i; j++) if (j%i == 0) { for (int k=0; k<j; k++) printf("*"); }} |
Solution: Consider the comments in the following function.
CPP
void function(int n){ int count = 0; // executes n times for (int i=0; i<n; i++) // executes O(n*n) times. for (int j=i; j< i*i; j++) if (j%i == 0) { // executes j times = O(n*n) times for (int k=0; k<j; k++) printf("*"); }} |
Time Complexity: O(n5)
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