Valid State of List - Google Top Interview Questions


Problem Statement :


Given a list of integers nums, return whether every number can be grouped using one of the following rules:

You can form contiguous pairs (a, a)

You can form contiguous triplets (a, a, a)

You can form contiguous triplets (a, a + 1, a + 2)

Constraints



n ≤ 100,000 where n is the length of nums.

Example 1

Input

nums = [7, 7, 3, 4, 5]

Output

True

Explanation

We can group [7, 7] together and [3, 4, 5] together.


Example 2

Input

nums = [1, 3, 2]

Output

False

Explanation

We can't group [1, 3, 2] together.


Solution :



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                        Solution in C++ :

bool solve(vector<int>& nums) {
    int n = nums.size();
    vector<int> dp(n + 1, 0);
    dp[0] = true;
    for (int i = 2; i <= n; i++) {
        if (nums[i - 1] == nums[i - 2]) {
            dp[i] = dp[i] | dp[i - 2];
        }
        if (i >= 3 && nums[i - 1] == nums[i - 2] && nums[i - 2] == nums[i - 3]) {
            dp[i] = dp[i] | dp[i - 3];
        }
        if (i >= 3 && nums[i - 3] + 2 == nums[i - 1] && nums[i - 2] + 1 == nums[i - 1]) {
            dp[i] = dp[i] | dp[i - 3];
        }
    }
    return dp[n];
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    private int n;
    private int[] a;
    private int[] dp;
    public boolean solve(int[] a) {
        this.a = a;
        n = a.length;
        dp = new int[n + 1];
        Arrays.fill(dp, -1);

        return solve(0);
    }

    private boolean solve(int i) {
        if (i == n)
            return true;

        if (dp[i] != -1) {
            return dp[i] == 1;
        }

        boolean ans = false;

        if (i + 1 < n && a[i + 1] == a[i]) {
            ans = ans || solve(i + 2);
        }

        if (i + 2 < n && a[i + 1] == a[i] && a[i + 2] == a[i]) {
            ans = ans || solve(i + 3);
        }

        if (i + 2 < n && a[i + 1] == a[i] + 1 && a[i + 2] == a[i] + 2) {
            ans = ans || solve(i + 3);
        }

        dp[i] = (ans ? 1 : 0);

        return ans;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums):
        #  setting the base case
        firstprev, secondprev, thirdprev = False, True, False
        for i in range(1, len(nums)):
            current = False

            #  first case checking pair (a, a)
            current |= secondprev if nums[i] == nums[i - 1] else False
            #  second case of checking triplets like (a,a,a)
            current |= thirdprev if i >= 2 and nums[i] == nums[i - 1] == nums[i - 2] else False
            # final case of checking the contiguous triples (a, a + 1, a + 2)
            current |= (
                thirdprev if i >= 2 and nums[i] == nums[i - 1] + 1 == nums[i - 2] + 2 else False
            )

            firstprev, secondprev, thirdprev = current, firstprev, secondprev

        return firstprev


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