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Reviewer's GuideThis PR adds C++ solutions for three LeetCode problems: sliding window approach for substring counting (problems 3297 and 3298) and a dynamic programming solution for "Delete and Earn" (problem 740), transforming the latter into a frequency-based DP. Class diagram for new Solution classes addedclassDiagram
class Solution_3297 {
+long long validSubstringCount(string word1, string word2)
}
class Solution_3298 {
+long long validSubstringCount(string word1, string word2)
}
class Solution_740 {
+int deleteAndEarn(vector<int>& nums)
}
Solution_3297 : validSubstringCount(word1, word2)
Solution_3298 : validSubstringCount(word1, word2)
Solution_740 : deleteAndEarn(nums)
Flow diagram for the dynamic programming approach in Delete and Earn (Problem 740)flowchart TD
A["Build frequency array f[i] from nums"] --> B["Find maxVal in nums"]
B --> C["Initialize dp[1] = f[1]"]
C --> D["For i = 2 to maxVal"]
D --> E["dp[i] = max(dp[i-1], dp[i-2] + f[i]*i)"]
E --> F["Return dp[maxVal]"]
File-Level Changes
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Intuition
The problem is similar to the "House Robber" problem but applied to numbers: if you take a number x, you cannot take x-1 or x+1.
The key insight is to transform the input array into a frequency array f[i], where f[i] counts how many times i appears. Now, instead of worrying about positions in the array, we only care about values. The maximum points we can earn for value i is f[i] * i, and we can use dynamic programming to decide whether to "take" or "skip" each number.
Approach
Build frequency array:
Count how many times each number appears in nums.
Keep track of the maximum number (maxVal) to limit DP size.
Dynamic Programming:
Let dp[i] be the maximum points we can earn considering numbers from 1 to i.
For each i ≥ 2, we have two options:
Skip i: dp[i] = dp[i-1]
Take i: dp[i] = dp[i-2] + f[i] * i (we skip i-1 because taking i deletes i-1)
Base case: dp[1] = f[1]
Return result:
dp[maxVal] is the maximum points we can earn.
Code Solution (C++)
class Solution {
public:
int deleteAndEarn(vector& nums) {
int m = 10001;
int n = nums.size();
int maxVal = 0;
vector f(m, 0);
vector dp(m + 1, 0);
};
Related Issues
Closes #254
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Summary by Sourcery
Add C++ solutions for LeetCode problems 3297, 3298, and 740
New Features: