leetcode

Author: ayoubzulfiqar

CheckIfThereIsAValidPartitionForTheArray/check_if_there_is_a_valid_partition_for_the_array.dart

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+/*
+
+-* 2369. Check if There is a Valid Partition For The Array *-
+
+You are given a 0-indexed integer array nums. You have to partition the array into one or more contiguous subarrays.
+
+We call a partition of the array valid if each of the obtained subarrays satisfies one of the following conditions:
+
+The subarray consists of exactly 2 equal elements. For example, the subarray [2,2] is good.
+The subarray consists of exactly 3 equal elements. For example, the subarray [4,4,4] is good.
+The subarray consists of exactly 3 consecutive increasing elements, that is, the difference between adjacent elements is 1. For example, the subarray [3,4,5] is good, but the subarray [1,3,5] is not.
+Return true if the array has at least one valid partition. Otherwise, return false.
+
+ 
+
+Example 1:
+
+Input: nums = [4,4,4,5,6]
+Output: true
+Explanation: The array can be partitioned into the subarrays [4,4] and [4,5,6].
+This partition is valid, so we return true.
+Example 2:
+
+Input: nums = [1,1,1,2]
+Output: false
+Explanation: There is no valid partition for this array.
+ 
+
+Constraints:
+
+2 <= nums.length <= 105
+1 <= nums[i] <= 106
+
+
+*/
+class Solution {
+  bool validPartition(List<int> nums) {
+    List<bool> vp = List.filled(nums.length + 1, true);
+    vp[0] = true;
+
+    if (nums[1] == nums[0]) {
+      vp[2] = true;
+    }
+    for (int i = 2; i < nums.length; i++) {
+      if (nums[i] == nums[i - 1]) {
+        vp[i + 1] = vp[i + 1] || vp[i - 1];
+      }
+      if (nums[i] == nums[i - 1] && nums[i] == nums[i - 2]) {
+        vp[i + 1] = vp[i + 1] || vp[i - 2];
+      }
+      if (nums[i] == nums[i - 1] + 1 && nums[i] == nums[i - 2] + 2) {
+        vp[i + 1] = vp[i + 1] || vp[i - 2];
+      }
+    }
+    return vp[nums.length];
+  }
+}

CheckIfThereIsAValidPartitionForTheArray/check_if_there_is_a_valid_partition_for_the_array.go

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+package main
+
+func validPartition(nums []int) bool {
+	n := len(nums)
+	vp := make([]bool, n+1)
+	vp[0] = true
+
+	if nums[1] == nums[0] {
+		vp[2] = true
+	}
+	for i := 2; i < n; i++ {
+		if nums[i] == nums[i-1] {
+			vp[i+1] = vp[i+1] || vp[i-1]
+		}
+		if nums[i] == nums[i-1] && nums[i] == nums[i-2] {
+			vp[i+1] = vp[i+1] || vp[i-2]
+		}
+		if nums[i] == nums[i-1]+1 && nums[i] == nums[i-2]+2 {
+			vp[i+1] = vp[i+1] || vp[i-2]
+		}
+	}
+	return vp[n]
+}