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39 lines
2.1 KiB
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<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of size <code>n</code> representing the cost of collecting different chocolates. The cost of collecting the chocolate at the index <code>i</code> is <code>nums[i]</code>. Each chocolate is of a different type, and initially, the chocolate at the index <code>i</code> is of <code>i<sup>th</sup></code> type.</p>
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<p>In one operation, you can do the following with an incurred <strong>cost</strong> of <code>x</code>:</p>
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<ul>
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<li>Simultaneously change the chocolate of <code>i<sup>th</sup></code> type to <code>((i + 1) mod n)<sup>th</sup></code> type for all chocolates.</li>
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</ul>
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<p>Return <em>the minimum cost to collect chocolates of all types, given that you can perform as many operations as you would like.</em></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> nums = [20,1,15], x = 5
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<strong>Output:</strong> 13
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<strong>Explanation:</strong> Initially, the chocolate types are [0,1,2]. We will buy the 1<sup>st</sup> type of chocolate at a cost of 1.
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Now, we will perform the operation at a cost of 5, and the types of chocolates will become [1,2,0]. We will buy the 2<sup>nd</sup><sup> </sup>type of chocolate at a cost of 1.
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Now, we will again perform the operation at a cost of 5, and the chocolate types will become [2,0,1]. We will buy the 0<sup>th </sup>type of chocolate at a cost of 1.
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Thus, the total cost will become (1 + 5 + 1 + 5 + 1) = 13. We can prove that this is optimal.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,3], x = 4
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<strong>Output:</strong> 6
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<strong>Explanation:</strong> We will collect all three types of chocolates at their own price without performing any operations. Therefore, the total cost is 1 + 2 + 3 = 6.
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</pre>
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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 <= nums.length <= 1000</code></li>
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<li><code>1 <= nums[i] <= 10<sup>9</sup></code></li>
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<li><code>1 <= x <= 10<sup>9</sup></code></li>
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</ul>
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