Add solution for Challenge 22 by Johrespi #654
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WalkthroughAdds two exported Go functions implementing a greedy coin-change algorithm: Changes
Estimated code review effort🎯 2 (Simple) | ⏱️ ~10 minutes
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Actionable comments posted: 1
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challenge-22/submissions/Johrespi/solution-template.go(1 hunks)
🔇 Additional comments (2)
challenge-22/submissions/Johrespi/solution-template.go (2)
7-27: LGTM!The main function provides clear demonstration of both functions with reasonable test cases.
31-52: Remove the algorithmic critique and focus only on valid code quality issues.This challenge explicitly requires a greedy approach for U.S. coin denominations [1, 5, 10, 25, 50], where it produces optimal results. The challenge documentation already acknowledges that greedy doesn't work for arbitrary denominations but explicitly states: "Your solution should implement the greedy approach, which always chooses the largest coin possible."
The following issues need addressing:
- Remove TODO comment on line 32
- Fix formatting on line 34: change
sum :=0tosum := 0The algorithm implementation itself aligns with the challenge requirements. Dynamic programming is not needed for this specific challenge.
Likely an incorrect or invalid review comment.
| func CoinCombination(amount int, denominations []int) map[int]int { | ||
| // TODO: Implement this function | ||
| c := make(map[int]int) | ||
| sum := 0 | ||
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| for i:= len(denominations) - 1; i >= 0; i--{ | ||
| for{ | ||
| if sum + denominations[i] <= amount{ | ||
| sum += denominations[i] | ||
| c[denominations[i]]++ | ||
| } else{ | ||
| break | ||
| } | ||
| } | ||
| } | ||
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| if sum != amount { | ||
| return map[int]int{} | ||
| } | ||
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| return c | ||
| } |
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🛠️ Refactor suggestion | 🟠 Major
Significant code duplication with MinCoins.
This function duplicates the core greedy logic from MinCoins (lines 36-45 vs lines 63-71). The same greedy algorithm issues and input validation concerns apply here.
Consider refactoring to eliminate duplication:
func CoinCombination(amount int, denominations []int) map[int]int {
combination := make(map[int]int)
remaining := amount
// Iterate from largest to smallest denomination
for i := len(denominations) - 1; i >= 0; i-- {
coin := denominations[i]
if coin <= remaining {
count := remaining / coin
if count > 0 {
combination[coin] = count
remaining -= count * coin
}
}
}
if remaining != 0 {
return map[int]int{}
}
return combination
}This refactored version:
- Uses division instead of nested loops for better efficiency
- Eliminates the duplicate loop logic
- More clearly expresses intent
- Removes the TODO comment (line 59)
Note: The greedy algorithm limitation mentioned in the MinCoins review also applies here.
🤖 Prompt for AI Agents
In challenge-22/submissions/Johrespi/solution-template.go around lines 58 to 79,
CoinCombination duplicates the greedy loop logic from MinCoins and uses
inefficient nested loops; replace the nested loop with a single-pass
division-based greedy approach: iterate denominations from largest to smallest,
compute count := remaining / coin, set combination[coin] = count if >0, subtract
count*coin from remaining, and after the loop return an empty map if remaining
!= 0; remove the TODO comment and consider extracting shared greedy logic into a
helper to eliminate duplication with MinCoins.
2 participants
Challenge 22 Solution
Submitted by: @Johrespi
Challenge: Challenge 22
Description
This PR contains my solution for Challenge 22.
Changes
challenge-22/submissions/Johrespi/solution-template.goTesting
Thank you for reviewing my submission! 🚀