**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 :

` ````
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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