The Longest Increasing Subsequence problem involves finding the longest subsequence of a given array of numbers, where the subsequence is in increasing order. However, in this variant of the problem, we have a specific cost function associated with each element of the sequence, which we need to take into account while finding the longest increasing subsequence.
To solve this problem using dynamic programming, we can use the following approach:
1. Create an array ‘dp‘ of size ‘n‘, where ‘n‘ is the length of the original sequence. ‘dp[i]‘ will store the length of the longest increasing subsequence ending at index ‘i‘.
2. Initialize ‘dp[i]‘ to ‘1‘ for all ‘i‘ in ‘[0, n-1]‘, as any element can form a subsequence of length 1.
3. For each index ‘i‘ in ‘[1, n-1]‘, iterate over all indices ‘j‘ in ‘[0, i-1]‘. If ‘a[j] < a[i]‘ and ‘cost[j] < cost[i]‘ (where ‘a‘ is the original sequence and ‘cost‘ is the array containing the cost function values), then update ‘dp[i]‘ as ‘max(dp[j] + 1, dp[i])‘. This means that if we can extend the longest increasing subsequence ending at index ‘j‘ with the element at index ‘i‘, and the cost of the subsequence doesn’t increase, then we should update the length of the longest increasing subsequence ending at index ‘i‘.
4. Finally, the solution to the problem is the maximum value in the ‘dp‘ array.
Here’s the Java code for this approach:
public int longestIncreasingSubsequenceWithCost(int[] a, int[] cost) {
int n = a.length;
int[] dp = new int[n];
// Initialize dp array to 1 as any element can form a subsequence of length 1.
Arrays.fill(dp, 1);
// Iterate over all indices i and j to update dp array.
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (a[j] < a[i] && cost[j] < cost[i]) {
dp[i] = Math.max(dp[j] + 1, dp[i]);
}
}
}
// Find the maximum value in the dp array, which is the length of the longest increasing subsequence with the given cost function.
int maxLength = 0;
for (int i = 0; i < n; i++) {
maxLength = Math.max(maxLength, dp[i]);
}
return maxLength;
}Let’s take an example to see how this algorithm works. Consider the sequence ‘[4, 2, 8, 5, 10]‘ and the cost function ‘[1, 4, 2, 3, 7]‘. We want to find the longest increasing subsequence with the given cost function.
The ‘dp‘ array will be initialized as ‘[1, 1, 1, 1, 1]‘. We then iterate over all possible ‘(i, j)‘ pairs to update the ‘dp‘ array:
- When `i = 1` and `j = 0`, we have `a[j] < a[i]` and `cost[j] < cost[i]`, so we update `dp[i]` as `max(dp[j] + 1, dp[i])`, which becomes `2`.
- When `i = 2` and `j = 0`, we have `a[j] < a[i]` and `cost[j] < cost[i]`, so we update `dp[i]` as `max(dp[j] + 1, dp[i])`, which becomes `2`.
- When `i = 2` and `j = 1`, we have `a[j] > a[i]`, so we skip this pair.
- When `i = 3` and `j = 0`, we have `a[j] < a[i]` and `cost[j] < cost[i]`, so we update `dp[i]` as `max(dp[j] + 1, dp[i])`, which becomes `2`.
- When `i = 3` and `j = 1`, we have `a[j] > a[i]`, so we skip this pair.
- When `i = 3` and `j = 2`, we have `a[j] < a[i]` and `cost[j] < cost[i]`, so we update `dp[i]` as `max(dp[j] + 1, dp[i])`, which becomes `3`.
- When `i = 4` and `j = 0`, we have `a[j] < a[i]` and `cost[j] < cost[i]`, so we update `dp[i]` as `max(dp[j] + 1, dp[i])`, which becomes `2`.
- When `i = 4` and `j = 1`, we have `a[j] > a[i]`, so we skip this pair.
- When `i = 4` and `j = 2`, we have `a[j] < a[i]` and `cost[j] < cost[i]`, so we update `dp[i]` as `max(dp[j] + 1, dp[i])`, which becomes `3`.
- When `i = 4` and `j = 3`, we have `a[j] < a[i]` and `cost[j] < cost[i]`, so we update `dp[i]` as `max(dp[j] + 1, dp[i])`, which becomes `4`.The final ‘dp‘ array becomes ‘[1, 2, 2, 3, 4]‘, and the solution to the problem is the maximum value in this array, which is ‘4‘. Therefore, the longest increasing subsequence with the given cost function is ‘[4, 8, 10]‘.
Overall, the time complexity of this algorithm is ‘O(n2)‘, as we have two nested loops to iterate over all possible ‘(i, j)‘ pairs. However, this solution can be optimized to ‘O(nlogn)‘ using binary search instead of nested loops.