Leetcode #1655: Distribute Repeating Integers
In this guide, we solve Leetcode #1655 Distribute Repeating Integers in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an array of n integers, nums, where there are at most 50 unique values in the array. You are also given an array of m customer order quantities, quantity, where quantity[i] is the amount of integers the ith customer ordered.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Array, Dynamic Programming, Backtracking, Bitmask
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: nums = [1,2,3,4], quantity = [2]
Output: false
Explanation: The 0th customer cannot be given two different integers.
Python Solution
class Solution:
def canDistribute(self, nums: List[int], quantity: List[int]) -> bool:
m = len(quantity)
s = [0] * (1 << m)
for i in range(1, 1 << m):
for j in range(m):
if i >> j & 1:
s[i] = s[i ^ (1 << j)] + quantity[j]
break
cnt = Counter(nums)
arr = list(cnt.values())
n = len(arr)
f = [[False] * (1 << m) for _ in range(n)]
for i in range(n):
f[i][0] = True
for i, x in enumerate(arr):
for j in range(1, 1 << m):
if i and f[i - 1][j]:
f[i][j] = True
continue
k = j
while k:
ok1 = j == k if i == 0 else f[i - 1][j ^ k]
ok2 = s[k] <= x
if ok1 and ok2:
f[i][j] = True
break
k = (k - 1) & j
return f[-1][-1]
Complexity
The time complexity is O(n * 3^m), and the space complexity is O(n * 2^m). The space complexity is O(n * 2^m).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.