Given a value `value`, if we want to make change for `value` cents, and we have infinite supply of each of coins = {S1, S2, .. , Sm} valued `coins`, how many ways can we make the change? The order of `coins` doesn't matter. For example, for `value` = 4 and `coins` = [1, 2, 3], there are four solutions: [1, 1, 1, 1], [1, 1, 2], [2, 2], [1, 3]. So output should be 4. For `value` = 10 and `coins` = [2, 5, 3, 6], there are five solutions: [2, 2, 2, 2, 2], [2, 2, 3, 3], [2, 2, 6], [2, 3, 5] and [5, 5]. So the output should be 5. Time complexity: O(n * m) where n is the `value` and m is the number of `coins` Space complexity: O(n) """ def count(coins, value):.
Given a value `value`, if we want to make change for `value` cents, and we have infinite supply of each of coins = {S1, S2, .. , Sm} valued `coins`, how many ways can we make the change? The order of `coins` doesn't matter. For example, for `value` = 4 and `coins` = [1, 2, 3], there are four solutions: [1, 1, 1, 1], [1, 1, 2], [2, 2], [1, 3]. So output should be 4. For `value` = 10 and `coins` = [2, 5, 3, 6], there are five solutions: [2, 2, 2, 2, 2], [2, 2, 3, 3], [2, 2, 6], [2, 3, 5] and [5, 5]. So the output should be 5. Time complexity: O(n * m) where n is the `value` and m is the number of `coins` Space complexity: O(n) """ def count(coins, value):.
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Question
Given a value `value`, if we want to make change for `value` cents, and we have infinite
supply of each of coins = {S1, S2, .. , Sm} valued `coins`, how many ways can we make the change?
The order of `coins` doesn't matter.
For example, for `value` = 4 and `coins` = [1, 2, 3], there are four solutions:
[1, 1, 1, 1], [1, 1, 2], [2, 2], [1, 3].
So output should be 4.
For `value` = 10 and `coins` = [2, 5, 3, 6], there are five solutions:
[2, 2, 2, 2, 2], [2, 2, 3, 3], [2, 2, 6], [2, 3, 5] and [5, 5].
So the output should be 5.
Time complexity: O(n * m) where n is the `value` and m is the number of `coins`
Space complexity: O(n)
"""
def count(coins, value):.
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