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1187. Make Array Strictly Increasing 👍

  • Time: $O(\texttt{sort})$
  • Space: $O(|\texttt{arr1}|)$
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class Solution {
 public:
  int makeArrayIncreasing(vector<int>& arr1, vector<int>& arr2) {
    // dp[i] := the minimum steps to reach i at previous round
    unordered_map<int, int> dp{{-1, 0}};

    ranges::sort(arr2);

    for (const int a : arr1) {
      unordered_map<int, int> newDp;
      for (const auto& [val, steps] : dp) {
        // It's possible to use the value in the arr1.
        if (a > val)
          newDp[a] = min(newDp.contains(a) ? newDp[a] : INT_MAX, steps);
        // Also try the value in the arr2.
        if (const auto it = ranges::upper_bound(arr2, val); it != arr2.cend())
          newDp[*it] =
              min(newDp.contains(*it) ? newDp[*it] : INT_MAX, steps + 1);
      }
      if (newDp.empty())
        return -1;
      dp = std::move(newDp);
    }

    int ans = INT_MAX;
    for (const auto& [_, steps] : dp)
      ans = min(ans, steps);
    return ans;
  }
};

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class Solution {
  public int makeArrayIncreasing(int[] arr1, int[] arr2) {
    // dp[i] := the minimum steps to reach i at previous round
    Map<Integer, Integer> dp = new HashMap<>();
    dp.put(-1, 0);

    Arrays.sort(arr2);

    for (final int a : arr1) {
      Map<Integer, Integer> newDp = new HashMap<>();
      for (final int val : dp.keySet()) {
        final int steps = dp.get(val);
        // It's possible to use the value in the arr1.
        if (a > val)
          newDp.put(a, Math.min(newDp.getOrDefault(a, Integer.MAX_VALUE), steps));
        // Also try the value in the arr2.
        final int i = firstGreater(arr2, val);
        if (i < arr2.length)
          newDp.put(arr2[i], Math.min(newDp.getOrDefault(arr2[i], Integer.MAX_VALUE), steps + 1));
      }
      if (newDp.isEmpty())
        return -1;
      dp = newDp;
    }

    return Collections.min(dp.values());
  }

  private int firstGreater(int[] A, int target) {
    final int i = Arrays.binarySearch(A, target + 1);
    return i < 0 ? -i - 1 : i;
  }
}

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class Solution:
  def makeArrayIncreasing(self, arr1: list[int], arr2: list[int]) -> int:
    # dp[i] := the minimum steps to reach i at previous round
    dp = {-1: 0}

    arr2.sort()

    for a in arr1:
      newDp = collections.defaultdict(lambda: math.inf)
      for val, steps in dp.items():
        # It's possible to use the value in the arr1.
        if a > val:
          newDp[a] = min(newDp[a], steps)
        # Also try the value in the arr2.
        i = bisect_right(arr2, val)
        if i < len(arr2):
          newDp[arr2[i]] = min(newDp[arr2[i]], steps + 1)
      if not newDp:
        return -1
      dp = newDp

    return min(dp.values())