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-2³¹ <= nums[i] <= 2³¹ - 1

23:T73c,# Intuition\nThe problem seems to involve finding the smallest missing positive integer in an array. The approach likely involves creating a hash array to mark the presence of numbers within a certain range and then iterating to find the first missing positive.\n\n# Approach\n1. **Build the Hash Array:**\n - Iterate through the given array `nums` and mark the presence of each positive integer in a hash array.\n - The hash array, `hashArray`, is a boolean array of size `n` (the size of the input array), initialized to `false`.\n\n2. **Find the First Missing Positive:**\n - Iterate through the hash array to find the first index where the value is `false`.\n - The missing positive integer is found by adding 1 to the index (since array indices are zero-based).\n\n3. **Handle Edge Case:**\n - If all numbers from 1 to n are present in the input array, return `n + 1`.\n\n# Complexity\n- Time complexity: $$O(n)$$\n - The algorithm iterates through the input array once to build the hash array and then iterates through the hash array to find the missing positive.\n\n- Space complexity: $$O(n)$$\n - Additional space is used for the hash array, which has a size equal to the input array. The space complexity is linear with respect to the input size.hash\n\n# Code\n```\nclass Solution {\npublic: \n int firstMissingPositive(vector& nums) {\n int n = nums.size();\n\n // Step 1: Build the hash array\n vector hashArray(n, false);\n for (int num : nums) {\n if (1 <= num && num <= n) {\n hashArray[num - 1] = true;\n }\n }\n\n // Step 2: Find the first missing positive\n for (int i = 0; i < n; ++i) {\n if (!hashArray[i]) {\n return i + 1;\n }\n }\n\n // If all numbers from 1 to n are present, return n+1\n return n + 1;\n}\n};\n```\n~ILYT24:T4e2,\n### Explanation:\n\n1. **Method Signature:**\n - **Input:** Array o

First Missing Positive

Problem Description

Problem Description

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