801 Minimum Swaps To Make Sequences Increasing

# 1. Question

We have two integer sequences`A`and`B`of the same non-zero length.
We are allowed to swap elements`A[i]`and`B[i]`. Note that both elements are in the same index position in their respective sequences.
At the end of some number of swaps,`A`and`B`are both strictly increasing. (A sequence is_strictly increasing_if and only if`A < A < A < ... < A[A.length - 1]`.)
Given A and B, return the minimum number of swaps to make both sequences strictly increasing. It is guaranteed that the given input always makes it possible.
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Example:
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Input: A = [1,3,5,4], B = [1,2,3,7]
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Output: 1
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Explanation:
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Swap A and B. Then the sequences are:
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A = [1, 3, 5, 7] and B = [1, 2, 3, 4]
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which are both strictly increasing.
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Note:
• `A, B`are arrays with the same length, and that length will be in the range`[1, 1000]`.
• `A[i], B[i]`are integer values in the range`[0, 2000]`.

# 2. Implementation

(1) DP

1.如果A[i - 1] < A[i] && B[i - 1] < B[i] && A[i - 1] < B[i] && B[i - 1] < A[i], 这个时候我们在第i个位置既可以swap也可以不swap
1. 1.
如果A[i - 1] < A[i] && B[i - 1] < B[i] && (A[i - 1] > B[i] || B[i - 1] > A[i]), 这个时候我们在第i个位置的操作和第i - 1个位置的操作必须一致
2. 2.
如果(A[i - 1] > A[i] || B[i - 1] > B[i]) && (A[i - 1] < B[i] && B[i - 1] < A[i]), 这个时候在第i个位置的操作和在第i - 1个位置的操作相反
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class Solution {
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public int minSwap(int[] A, int[] B) {
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int n = A.length;
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int[][] dp = new int[n];
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// dp[i] means we don't swap at ith position
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// dp[i] means we swap at ith position
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dp = 0;
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dp = 1;
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for (int i = 1; i < n; i++) {
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// Case 1: In this case we can either swap or stay the same at ith poition
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if (A[i - 1] < A[i] && B[i - 1] < B[i] && A[i - 1] < B[i] && B[i - 1] < A[i]) {
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dp[i] = Math.min(dp[i - 1], dp[i - 1]);
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dp[i] = Math.min(dp[i - 1], dp[i - 1]) + 1;
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}
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// Case 2: If we swap at (i - 1)th position, we should swap at ith position
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// If we don't sway at (i - 1)th position, we keep the same operation at ith position
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else if (A[i - 1] < A[i] && B[i - 1] < B[i]) {
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dp[i] = dp[i - 1];
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dp[i] = dp[i - 1] + 1;
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}
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// Case 3: If we swap at (i - 1)th position, we don't swap at ith position
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// If we don't swap at (i - 1)th position, we swap at ith position
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else {
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dp[i] = dp[i - 1];
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dp[i] = dp[i - 1] + 1;
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}
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}
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return Math.min(dp[n - 1], dp[n - 1]);
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}
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}
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# 3. Time & Space Complexity

DP: 时间复杂度O(n), 空间复杂度O(n)