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Problem solution · Java

Sum of Consecutive Subsequences

Sum of Consecutive Subsequences: a Java solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
57 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

Hash-based lookup

For Sum of Consecutive Subsequences, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 57 lines of Java from the credited upstream file 3299.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 4 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 codeSum of Consecutive Subsequences · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int getSum(int[] nums) {    final long increasingSequenceSum = getSequenceSum(nums, 1);    final long decreasingSequenceSum = getSequenceSum(nums, -1);    final long arraySum = getArraySum(nums);    return (int) ((increasingSequenceSum + decreasingSequenceSum - arraySum + MOD) % MOD);  }   private static final int MOD = 1_000_000_007;   // Returns the sum of all sequences in the array that are in consecutive  // increasing order if `direction` is 1, or in consecutive decreasing order if  // `direction` is -1.  private long getSequenceSum(int[] nums, int direction) {    int n = nums.length;    long sequenceSum = 0;    // {num: the number of subsequences ending in `num` so far}    Map<Integer, Integer> prefixCount = new HashMap<>();    // {num: the number of subsequences starting from `num` so far}    Map<Integer, Integer> suffixCount = new HashMap<>();    // prefixSubseqs[i] := the number of subsequences ending in nums[i]    int[] prefixSubseqs = new int[n];    // suffixSubseqs[i] := the number of subsequences starting from nums[i]    int[] suffixSubseqs = new int[n];     for (int i = 0; i < n; ++i) {      final int prevNum = nums[i] - direction;      final int subseqsCount = prefixCount.getOrDefault(prevNum, 0) + 1;      prefixSubseqs[i] = subseqsCount;      prefixCount.merge(nums[i], subseqsCount, Integer::sum);      prefixCount.put(nums[i], prefixCount.get(nums[i]) % MOD);    }     for (int i = n - 1; i >= 0; --i) {      int nextNum = nums[i] + direction;      int subseqsCount = suffixCount.getOrDefault(nextNum, 0) + 1;      suffixSubseqs[i] = subseqsCount;      suffixCount.merge(nums[i], subseqsCount, Integer::sum);      suffixCount.put(nums[i], suffixCount.get(nums[i]) % MOD);    }     for (int i = 0; i < n; ++i) {      sequenceSum += (long) nums[i] * prefixSubseqs[i] % MOD * suffixSubseqs[i];      sequenceSum %= MOD;    }     return sequenceSum;  }   private int getArraySum(int[] nums) {    int arraySum = 0;    for (int num : nums)      arraySum = (arraySum + num) % MOD;    return arraySum;  }}

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