CodeSignal ---- almostIncreasingSequence

Given a sequence of integers as an array, determine whether it is possible to obtain a strictly increasing sequence by removing no more than one element from the array.

Note: sequence a0, a1, ..., an is considered to be a strictly increasing if a0 < a1 < ... < an. Sequence containing only one element is also considered to be strictly increasing.

Example

For sequence = [1, 3, 2, 1], the output should be solution(sequence) = false. There is no one element in this array that can be removed in order to get a strictly increasing sequence. For sequence = [1, 3, 2], the output should be solution(sequence) = true. You can remove 3 from the array to get the strictly increasing sequence [1, 2]. Alternately, you can remove 2 to get the strictly increasing sequence [1, 3].

Input/Output

[execution time limit] 3 seconds (java) [input] array.integer sequence Guaranteed constraints: 2 ≤ sequence.length ≤ 105, -105 ≤ sequence[i] ≤ 105. [output] boolean Return true if it is possible to remove one element from the array in order to get a strictly inc public class almostIncreasingSequence { public static void main(String args[]) { } private int FindDecreasingElement(int[] sequence) { for (int i = 1; i < sequence.length; i ++) { if (sequence[i - 1] >= sequence[i]) return i; } return sequence.length; } private boolean almostIncreasingSequence(int[] sequence) { int increasingCount = findDecreasingElement(sequence); if (increasingCount == sequence.length) { return true; } if (increasingCount == 1 || sequence[increasingCount - 2] < sequence[increasingCount]) { // if increasingCount == 1 or very big so we remove that return FindDecreasingElement(sequence, increasingCount) == sequence.length; } if (sequence[increasingCount - 1] < sequence[increasingCount + 1]) { // the increasingCount is a lot small, so remove that there return findDecreasingElement(sequence, increasingCount + 1) == sequence.length; } return false; } } reasing sequence, otherwise return false.