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Problem solution · Java

Find Products of Elements of Big Array

Find Products of Elements of Big Array: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
82 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

Direct simulation

For Find Products of Elements of Big Array, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 82 lines of Java from the credited upstream file 3145.java.
  • The implementation visibly relies on sequence storage.
  • 5 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeFind Products of Elements of Big Array · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int[] findProductsOfElements(long[][] queries) {    int[] ans = new int[queries.length];     for (int i = 0; i < queries.length; ++i) {      final long a = queries[i][0];      final long b = queries[i][1];      final int mod = (int) queries[i][2];      ans[i] = (int) modPow(2,                            sumPowersFirstKBigNums(b + 1) - //                                sumPowersFirstKBigNums(a),                            mod);    }     return ans;  }   // Returns the sum of powers of the first k numbers in `big_nums`.  private long sumPowersFirstKBigNums(long k) {    final long num = firstNumberHavingSumBitsTillGreaterThan(k);    long sumPowers = sumPowersTill(num - 1);    long remainingCount = k - sumBitsTill(num - 1);    for (int power = 0; power < bitLength(num); ++power) {      if ((num >> power & 1) == 1) {        sumPowers += power;        --remainingCount;        if (remainingCount == 0)          break;      }    }    return sumPowers;  }   // Returns the first number in [1, k] that has sumBitsTill(num) >= k.  private long firstNumberHavingSumBitsTillGreaterThan(long k) {    long l = 1;    long r = k;    while (l < r) {      final long m = (l + r) / 2;      if (sumBitsTill(m) < k)        l = m + 1;      else        r = m;    }    return l;  }   // Returns sum(i.bit_count()), where 1 <= i <= x.  private long sumBitsTill(long x) {    long sumBits = 0;    for (long powerOfTwo = 1; powerOfTwo <= x; powerOfTwo *= 2) {      sumBits += (x / (2 * powerOfTwo)) * powerOfTwo;      sumBits += Math.max(0, x % (2 * powerOfTwo) + 1 - powerOfTwo);    }    return sumBits;  }   // Returns sum(all powers of i), where 1 <= i <= x.  private long sumPowersTill(long x) {    long sumPowers = 0;    long powerOfTwo = 1;    for (int power = 0; power < bitLength(x); ++power) {      sumPowers += (x / (2 * powerOfTwo)) * powerOfTwo * power;      sumPowers += Math.max(0, x % (2 * powerOfTwo) + 1 - powerOfTwo) * power;      powerOfTwo *= 2;    }    return sumPowers;  }   private long modPow(long x, long n, int mod) {    if (n == 0)      return 1 % mod;    if (n % 2 == 1)      return x * modPow(x % mod, (n - 1), mod) % mod;    return modPow(x * x % mod, (n / 2), mod) % mod;  }   private int bitLength(long x) {    return Long.SIZE - Long.numberOfLeadingZeros(x);  }}

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