Given n elements, write a program that prints the longest increasing subsequence whose adjacent element difference is one.

Examples:

Input : a[] = {3, 10, 3, 11, 4, 5, 6, 7, 8, 12}

Output : 3 4 5 6 7 8

Explanation: 3, 4, 5, 6, 7, 8 is the longest increasing subsequence whose adjacent element differs by one.Input : a[] = {6, 7, 8, 3, 4, 5, 9, 10}

Output : 6 7 8 9 10

Explanation: 6, 7, 8, 9, 10 is the longest increasing subsequence

We have discussed how to find length of Longest Increasing consecutive subsequence. To print the subsequence, we store index of last element. Then we print consecutive elements ending with last element.

Given below is the implementation of the above approach:

// CPP program to find length of the // longest increasing subsequence // whose adjacent element differ by 1 #include <bits/stdc++.h> using namespace std; // function that returns the length of the // longest increasing subsequence // whose adjacent element differ by 1 void longestSubsequence(int a[], int n) { // stores the index of elements unordered_map<int, int> mp; // stores the length of the longest // subsequence that ends with a[i] int dp[n]; memset(dp, 0, sizeof(dp)); int maximum = INT_MIN; // iterate for all element int index = -1; for (int i = 0; i < n; i++) { // if a[i]-1 is present before i-th index if (mp.find(a[i] - 1) != mp.end()) { // last index of a[i]-1 int lastIndex = mp[a[i] - 1] - 1; // relation dp[i] = 1 + dp[lastIndex]; } else dp[i] = 1; // stores the index as 1-index as we need to // check for occurrence, hence 0-th index // will not be possible to check mp[a[i]] = i + 1; // stores the longest length if (maximum < dp[i]) { maximum = dp[i]; index = i; } } // We know last element of sequence is // a[index]. We also know that length // of subsequence is "maximum". So We // print these many consecutive elements // starting from "a[index] - maximum + 1" // to a[index]. for (int curr = a[index] - maximum + 1; curr <= a[index]; curr++) cout << curr << " "; } // Driver Code int main() { int a[] = { 3, 10, 3, 11, 4, 5, 6, 7, 8, 12 }; int n = sizeof(a) / sizeof(a[0]); longestSubsequence(a, n); return 0; }

Output:

3 4 5 6 7 8

Time Complexity : O(n)

Auxiliary Space : O(n)

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