Added 1608-special-array-with-x-elements-greater-than-or-equal-x

dsa-solutions/lc-solutions/1600-1699/1608-special-array-with-x-elements-greater-than-or-equal-x.md

@@ -0,0 +1,127 @@
+---
+id: special-array-with-x-elements-greater-than-or-equal-x
+title: Special Array with X Elements Greater Than or Equal X
+level: medium
+sidebar_label: Special Array with X Elements Greater Than or Equal X
+tags:
+  - Array
+  - Sorting
+  - Counting
+  - Java
+description: "This document provides solutions for the Special Array with X Elements Greater Than or Equal X problem."
+---
+
+## Problem Statement
+
+You are given an array `nums` of non-negative integers. `nums` is considered special if there exists a number `x` such that there are exactly `x` numbers in `nums` that are greater than or equal to `x`.
+
+Notice that `x` does not have to be an element in `nums`.
+
+Return `x` if the array is special, otherwise, return `-1`. It can be proven that if `nums` is special, the value for `x` is unique.
+
+**Example 1:**
+
+Input: `nums = [3,5]`
+
+Output: `2`
+
+Explanation: There are 2 values (3 and 5) that are greater than or equal to 2.
+
+**Example 2:**
+
+Input: `nums = [0,0]`
+
+Output: `-1`
+
+Explanation: No numbers fit the criteria for `x`.
+- If `x = 0`, there should be 0 numbers >= `x`, but there are 2.
+- If `x = 1`, there should be 1 number >= `x`, but there are 0.
+- If `x = 2`, there should be 2 numbers >= `x`, but there are 0.
+- `x` cannot be greater since there are only 2 numbers in `nums`.
+
+**Example 3:**
+
+Input: `nums = [0,4,3,0,4]`
+
+Output: `3`
+
+Explanation: There are 3 values that are greater than or equal to 3.
+
+**Constraints:**
+
+- `1 <= nums.length <= 100`
+- `0 <= nums[i] <= 1000`
+
+## Solutions
+
+### Intuition
+
+To solve this problem, we need to determine if there exists a number `x` such that there are exactly `x` elements in the array that are greater than or equal to `x`. We can achieve this by counting the number of elements for each possible value and checking if the condition holds.
+
+### Approach
+
+1. **Counting Frequencies:**
+   - Create an array `intArr` to count the occurrences of each element in `nums`.
+
+2. **Cumulative Sum:**
+   - Calculate the cumulative sum from the highest value to the lowest to determine how many elements are greater than or equal to each possible value.
+
+3. **Check Condition:**
+   - Iterate through the possible values and check if there exists a value `x` where the number of elements greater than or equal to `x` is exactly `x`.
+
+### Java 
+
+```java
+class Solution {
+    public int specialArray(int[] nums) {
+        int val = 0;
+        int len = nums.length;
+        int[] intArr = new int[1001];
+
+        for(int i = 0; i < len; i++) {
+            intArr[nums[i]]++;
+        }
+
+        for(int i = 1000; i >= 0; i--) {
+            intArr[i] += val;
+            val = intArr[i];
+        }
+
+        val = -1;
+        for(int i = 0; i < 1000; i++) {
+            if(intArr[i] == i) {
+                val = i;
+            }
+        }
+
+        return val;
+    }
+
+}
+```
+
+### Python
+
+```Python
+class Solution:
+    def specialArray(self, nums: List[int]) -> int:
+        n = len(nums)
+        intArr = [0] * 1001
+
+        for num in nums:
+            if num <= 1000:
+                intArr[num] += 1
+
+        val = 0
+        for i in range(1000, -1, -1):
+            intArr[i] += val
+            val = intArr[i]
+
+        val = -1
+        for i in range(1001):
+            if intArr[i] == i:
+                val = i
+                break
+
+        return val
+```