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Longest Increasing subarray with one change allowed

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Given an array, find the length of the longest increasing subarray (contiguous elements) such that it is possible to change at most one number (change one number to any integer you want) from the sequence to make the sequence strictly increasing.

Examples: 

Input  : 6
         7 2 3 1 5 10 
Output : 5
Explanation : 
Here, we can choose subarray 2, 3, 1, 5, 10 
and by changing its 3rd element (that is 1) 
to 4, it will become increasing sequence.

Input  : 2
         10 10
Output : 2
Explanation : 
Here, we can choose subarray 10, 10 and by
changing its 2nd element (that is 10) to 11,
it will become increasing sequence.

Approach : 

Step 1: We first compute the longest increasing subarray ending at an index for every index in the given array. We store these values in l[]. 
Step 2: Then calculate the longest increasing subarray starting at an index for every index in the given array. We store these values in r[]. 
Step 3: Update the answer ans = max ( ans, l[i-1] + r[i+1] + 1), when a[i-1] + 1 < a[i+1].

Below is the implementation of the above idea: 

C++





Java





Python3





C#





Javascript




<script>
 
// Javascript program to find
// longest increasing subarray
// with one change allowed.
     
    // Function to find length of
    // subsequence
    function seg(a,n)
    {
        let l = new Array(n);
        let r = new Array(n + 1);
        let ans = 0;
  
        // calculating the l array.
        l[0] = 1;
        for (let i = 1; i < n; i++)
        {
            if (a[i] > a[i - 1])
            {
                l[i] = l[i - 1] + 1;
                ans = Math.max(ans, l[i]);
            }
            else
                l[i] = 1;
        }
        if (ans != n)
            ++ans;
  
        // calculating the r array.
        r[n] = 0;
        for (let i = n - 1; i >= 0; i--)
        {
            if (a[i - 1] < a[i])
                r[i] = r[i + 1] + 1;
            else
                r[i] = 1;
        }
  
        // updating the answer.
        for (let i = n - 2; i > 0; i--)
        {
            if (a[i + 1] - a[i - 1] > 1)
                ans = Math.max(ans,
                            l[i - 1] + r[i + 1] + 1);
        }
        return Math.max(ans, r[0]);
    }
     
    // Driver code.
    let a=[9, 4, 5, 1, 13 ];
    let n = a.length;
     
    document.write(seg(a, n));
     
    // This code is contributed by rag2127
     
</script>


Output

4


Last Updated : 22 Jul, 2022
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