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admin 11月 14, 2020 0

Pazzle

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.

Example:
For sequence = [1, 3, 2, 1], the output should be
almostIncreasingSequence(sequence) = false.

For sequence = [1, 3, 2], the output should be
almostIncreasingSequence(sequence) = true

Ref - https://app.codesignal.com/arcade/intro/level-2/2mxbGwLzvkTCKAJMG

Test


public void () {
    boolean act = almostIncreasingSequence(new int[]{1, 3, 2, 1});
    assertFalse(act);
}


public void test_case_2() {
    boolean act = almostIncreasingSequence(new int[]{1, 3, 2});
    assertTrue(act);
}


public void test_case_3() {
    boolean act = almostIncreasingSequence(new int[]{10, 1, 2, 3, 4, 5});
    assertTrue(act);
}
    

public void test_case_4() {
    boolean act = almostIncreasingSequence(new int[]{1, 2, 1, 2});
    assertFalse(act);
}


public void test_case_5() {
    boolean act = almostIncreasingSequence(new int[]{1, 3});
    assertTrue(act);
}

Solution

一開始的想法直覺是想到排序後做比較, diff > 2 以上 return false, 但不太對,
像 [10, 1, 2, 3, 4, 5] 排序後, diff 絕對大於 2, 且也是屬於 almost incrasing seq,
最後回歸基本迴圈 foreach 比較

boolean almostIncreasingSequence(int[] seq) {
    int countNoSeq = 0;
    
    for (int i = 0; i < seq.length - 1; i++) {
        if (seq[i] >= seq[i + 1]) {
            countNoSeq++;
        }
    }
    // compare with after next element, because removed an element
    for (int i = 0; i < seq.length - 2; i++) {
        if (seq[i] >= seq[i + 2]) {
            countNoSeq++;
        }
    }

    if (countNoSeq > 2) {
        return false;
    }

    return true;
}

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