leetcode new problems

leetcode/lc3637/betterTrionicArray.cpp

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+class Solution {
+public:
+    bool isTrionic(vector<int>& nums) {
+        int n = nums.size(), i = 1;
+        while (i < n && nums[i - 1] < nums[i]) {
+            i++;
+        }
+        int p = i - 1;
+        while (i < n && nums[i - 1] > nums[i]) {
+            i++;
+        }
+        int q = i - 1;
+        while (i < n && nums[i - 1] < nums[i]) {
+            i++;
+        }
+        int flag = i - 1;
+        return (p != 0) && (q != p) && (flag == n - 1 && flag != q);
+    }
+};
+
+// Uses single pointer, and checks for all 3 required segments one after one.a

leetcode/lc3637/trionicArray.cpp

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+class Solution {
+    public : 
+        bool isTrionic(vector<int> &nums){
+            int n =nums.size();
+            if(n<3) return false;
+            int p = 0;
+            int q = n-1;
+            while(p+1 <n && nums[p] < nums[p+1]) p++;
+            while(q-1>=0 && nums[q] > nums[q-1]) q--;
+            if(p==0 || q==n-1) return false;
+            while(p+1<=q){
+                if(nums[p] > nums[p+1]) p++;
+                else return false;
+            }
+            return p==q;
+        }
+};
+
+// Uses two pointers, TC : o(n) , SC : o(1)
+// finds valid p and q if they exist, adn checks for strictly decreasing between p...q

leetcode/lc3637/trionicSegmentCount.cpp

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+class Solution {
+public:
+    bool isTrionic(vector<int>& nums) {
+        int n = nums.size();
+        if (nums[0] >= nums[1]) {
+            return false;
+        }
+        int count = 1;
+        for (int i = 2; i < n; i++) {
+            if (nums[i - 1] == nums[i]) {
+                return false;
+            }
+            if ((nums[i - 2] - nums[i - 1]) * (nums[i - 1] - nums[i]) < 0) {
+                count++;
+            }
+        }
+        return count == 3;
+    }
+};
+// count denotes the number of monotonic segments and 3 elements are considered at a time, and checked for sign flip, if theres sign flip then segment has changed, starting from increasing segment, if count ==3 then it has three monotonic segments and hence return true