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Problem solution · Java

Minimum Relative Loss After Buying Chocolates

Minimum Relative Loss After Buying Chocolates: a Java solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Binary search
Source
walkccc LeetCode Solutions
Length
74 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Binary search

For Minimum Relative Loss After Buying Chocolates, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 74 lines of Java from the credited upstream file 2819.java.
  • The implementation visibly relies on sequence storage.
  • 3 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Relative Loss After Buying Chocolates · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  long[] minimumRelativeLosses(int[] prices, int[][] queries) {    final int n = prices.length;    long[] ans = new long[queries.length];    long[] prefix = new long[prices.length + 1];     Arrays.sort(prices);     for (int i = 0; i < prices.length; ++i)      prefix[i + 1] = prefix[i] + prices[i];     for (int i = 0; i < queries.length; ++i) {      final int k = queries[i][0];      final int m = queries[i][1];      final int countFront = getCountFront(k, m, prices);      final int countBack = m - countFront;      ans[i] = getRelativeLoss(countFront, countBack, k, prefix);    }     return ans;  }   // Returns `countFront` for query (k, m) s.t. picking the first `countFront`  // and the last `m - countFront` chocolates is optimal.  //  // Define loss[i] := the relative loss of picking `prices[i]`.  // 1. For prices[i] <= k, Bob pays prices[i] while Alice pays 0.  //    Thus, loss[i] = prices[i] - 0 = prices[i].  // 2. For prices[i] > k, Bob pays k while Alice pays prices[i] - k.  //    Thus, loss[i] = k - (prices[i] - k) = 2 * k - prices[i].  // By observation, we deduce that it is always better to pick from the front  // or the back since loss[i] is increasing for 1. and is decreasing for 2.  //  // Assume that picking `left` chocolates from the left and `right = m - left`  // chocolates from the right is optimal. Therefore, we are selecting  // chocolates from `prices[0..left - 1]` and `prices[n - right..n - 1]`.  //  // To determine the optimal `left` in each iteration, we simply compare  // `loss[left]` with `loss[n - right]`; if `loss[left] < loss[n - right]`,  // it's worth increasing `left`.  private int getCountFront(int k, int m, int[] prices) {    final int n = prices.length;    final int countNoGreaterThanK = firstGreater(prices, k);    int l = 0;    int r = Math.min(countNoGreaterThanK, m);     while (l < r) {      final int mid = (l + r) / 2;      final int right = m - mid;      // Picking prices[mid] is better than picking prices[n - right].      if (prices[mid] < 2L * k - prices[n - right])        l = mid + 1;      else        r = mid;    }     return l;  }   private int firstGreater(int[] arr, int target) {    final int i = Arrays.binarySearch(arr, target + 1);    return i < 0 ? -i - 1 : i;  }   // Returns the relative loss of picking `countFront` and `countBack`  // chocolates.  private long getRelativeLoss(int countFront, int countBack, int k, long[] prefix) {    final int n = prefix.length - 1;    final long lossFront = prefix[countFront];    final long lossBack = 2L * k * countBack - (prefix[n] - prefix[n - countBack]);    return lossFront + lossBack;  }}

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