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<p>Given a <strong>0-indexed</strong> array of strings <code>words</code> where <code>words[i]</code> is either a positive integer represented as a string or the string <code>"prev"</code>.</p>
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<p>Given an integer array <code>nums</code> where <code>nums[i]</code> is either a positive integer or <code>-1</code>. We need to find for each <code>-1</code> the respective positive integer, which we call the last visited integer.</p>
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<p>Start iterating from the beginning of the array; for every <code>"prev"</code> string seen in <code>words</code>, find the <strong>last visited integer</strong> in <code>words</code> which is defined as follows:</p>
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<p>To achieve this goal, let's define two empty arrays: <code>seen</code> and <code>ans</code>.</p>
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<p>Start iterating from the beginning of the array <code>nums</code>.</p>
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<ul>
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<li>Let <code>k</code> be the number of consecutive <code>"prev"</code> strings seen so far (containing the current string). Let <code>nums</code> be the <strong>0-indexed </strong>array of <strong>integers</strong> seen so far and <code>nums_reverse</code> be the reverse of <code>nums</code>, then the integer at <code>(k - 1)<sup>th</sup></code> index of <code>nums_reverse</code> will be the <strong>last visited integer</strong> for this <code>"prev"</code>.</li>
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<li>If <code>k</code> is <strong>greater</strong> than the total visited integers, then the last visited integer will be <code>-1</code>.</li>
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<li>If a positive integer is encountered, prepend it to the <strong>front</strong> of <code>seen</code>.</li>
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<li>If <code>-1</code> is encountered, let <code>k</code> be the number of <strong>consecutive</strong> <code>-1</code>s seen so far (including the current <code>-1</code>),
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<ul>
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<li>If <code>k</code> is less than or equal to the length of <code>seen</code>, append the <code>k</code>-th element of <code>seen</code> to <code>ans</code>.</li>
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<li>If <code>k</code> is strictly greater than the length of <code>seen</code>, append <code>-1</code> to <code>ans</code>.</li>
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</ul>
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</li>
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</ul>
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<p>Return <em>an integer array containing the last visited integers.</em></p>
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<p>Return the array<em> </em><code>ans</code>.</p>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> words = ["1","2","prev","prev","prev"]
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<strong>Output:</strong> [2,1,-1]
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<strong>Explanation:</strong>
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For "prev" at index = 2, last visited integer will be 2 as here the number of consecutive "prev" strings is 1, and in the array reverse_nums, 2 will be the first element.
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For "prev" at index = 3, last visited integer will be 1 as there are a total of two consecutive "prev" strings including this "prev" which are visited, and 1 is the second last visited integer.
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For "prev" at index = 4, last visited integer will be -1 as there are a total of three consecutive "prev" strings including this "prev" which are visited, but the total number of integers visited is two.
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</pre>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">nums = [1,2,-1,-1,-1]</span></p>
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<p><strong>Output:</strong> <span class="example-io">[2,1,-1]</span></p>
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<p><strong>Explanation:</strong></p>
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<p>Start with <code>seen = []</code> and <code>ans = []</code>.</p>
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<ol>
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<li>Process <code>nums[0]</code>: The first element in nums is <code>1</code>. We prepend it to the front of <code>seen</code>. Now, <code>seen == [1]</code>.</li>
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<li>Process <code>nums[1]</code>: The next element is <code>2</code>. We prepend it to the front of <code>seen</code>. Now, <code>seen == [2, 1]</code>.</li>
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<li>Process <code>nums[2]</code>: The next element is <code>-1</code>. This is the first occurrence of <code>-1</code>, so <code>k == 1</code>. We look for the first element in seen. We append <code>2</code> to <code>ans</code>. Now, <code>ans == [2]</code>.</li>
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<li>Process <code>nums[3]</code>: Another <code>-1</code>. This is the second consecutive <code>-1</code>, so <code>k == 2</code>. The second element in <code>seen</code> is <code>1</code>, so we append <code>1</code> to <code>ans</code>. Now, <code>ans == [2, 1]</code>.</li>
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<li>Process <code>nums[4]</code>: Another <code>-1</code>, the third in a row, making <code>k = 3</code>. However, <code>seen</code> only has two elements (<code>[2, 1]</code>). Since <code>k</code> is greater than the number of elements in <code>seen</code>, we append <code>-1</code> to <code>ans</code>. Finally, <code>ans == [2, 1, -1]</code>.</li>
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</ol>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> words = ["1","prev","2","prev","prev"]
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<strong>Output:</strong> [1,2,1]
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<strong>Explanation:</strong>
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For "prev" at index = 1, last visited integer will be 1.
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For "prev" at index = 3, last visited integer will be 2.
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For "prev" at index = 4, last visited integer will be 1 as there are a total of two consecutive "prev" strings including this "prev" which are visited, and 1 is the second last visited integer.
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</pre>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">nums = [1,-1,2,-1,-1]</span></p>
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<p><strong>Output:</strong><span class="example-io"> [1,2,1]</span></p>
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<p><strong>Explanation:</strong></p>
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<p>Start with <code>seen = []</code> and <code>ans = []</code>.</p>
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<ol>
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<li>Process <code>nums[0]</code>: The first element in nums is <code>1</code>. We prepend it to the front of <code>seen</code>. Now, <code>seen == [1]</code>.</li>
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<li>Process <code>nums[1]</code>: The next element is <code>-1</code>. This is the first occurrence of <code>-1</code>, so <code>k == 1</code>. We look for the first element in <code>seen</code>, which is <code>1</code>. Append <code>1</code> to <code>ans</code>. Now, <code>ans == [1]</code>.</li>
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<li>Process <code>nums[2]</code>: The next element is <code>2</code>. Prepend this to the front of <code>seen</code>. Now, <code>seen == [2, 1]</code>.</li>
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<li>Process <code>nums[3]</code>: The next element is <code>-1</code>. This <code>-1</code> is not consecutive to the first <code>-1</code> since <code>2</code> was in between. Thus, <code>k</code> resets to <code>1</code>. The first element in <code>seen</code> is <code>2</code>, so append <code>2</code> to <code>ans</code>. Now, <code>ans == [1, 2]</code>.</li>
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<li>Process <code>nums[4]</code>: Another <code>-1</code>. This is consecutive to the previous <code>-1</code>, so <code>k == 2</code>. The second element in <code>seen</code> is <code>1</code>, append <code>1</code> to <code>ans</code>. Finally, <code>ans == [1, 2, 1]</code>.</li>
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</ol>
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</div>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 <= words.length <= 100</code></li>
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<li><code>words[i] == "prev"</code> or <code>1 <= int(words[i]) <= 100</code></li>
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<li><code>1 <= nums.length <= 100</code></li>
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<li><code>nums[i] == -1</code> or <code>1 <= nums[i] <= 100</code></li>
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</ul>
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