Tome Voyage Partition 15 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing tome and voyage metrics, construct an optimal algorithm to evaluate and compute the target partition value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Voyage Partition 15"
WHY DOES IT MATTER?
Backtracking provides a systematic way to explore combinatorial spaces while discarding impossible paths early.
OPTIMIZATION CHALLENGE
The key is reducing exponential blow‑up by pruning and state reuse to achieve practical runtimes.
REAL-WORLD CONNECTION
It mirrors resource allocation in logistics where each shipment must fit capacity constraints without exhaustive trial.
Always sort inputs and track remaining capacity; this simple ordering yields the biggest pruning gains.
COMPLEXITY AT A GLANCE
O(2^n) in the worst case, often much lower with pruningO(n) recursion stack plus memoizationCore Theory — Why This Approach?
Backtracking solves partition‑type problems by exploring a decision tree where each element can be assigned to one of several groups, pruning branches that violate constraints early. For the Tome Voyage Partition, the naive recursion enumerates all 2^n assignments, quickly exhausting time limits as n grows beyond 20.
The optimal paradigm augments plain recursion with stateful pruning: maintain the current sum of the target partition, reject any branch where the sum exceeds the required value, and use memoization or ordering heuristics (e.g., sorting descending) to cut off symmetric states. This reduces the effective search space dramatically, turning an exponential blow‑up into a tractable exploration for typical interview constraints.
Interview Questions on This Problem
Q1How does backtracking differ from brute‑force recursion in handling partition problems?
Backtracking adds early exit conditions that discard infeasible partial solutions, while brute‑force explores every combination regardless of constraints. This pruning dramatically lowers the number of recursive calls.
Q2Why is sorting the input array often a beneficial preprocessing step for backtracking partition solutions?
Sorting descending places larger elements first, causing constraint violations earlier and enabling quicker pruning. It also helps avoid duplicate work by handling equal values in a deterministic order.
Q3What is the role of memoization in a backtracking solution for the Tome Voyage Partition?
Memoization caches results of sub‑problems defined by the current index and remaining target sum, preventing re‑evaluation of identical states. This transforms overlapping recursive calls into O(n·target) lookups.
Examples
Input
[150, 75, 25, 50, 100, 125, 175, 200, 225, 250]
Output
150
Explanation: Step-by-step: Given the input array [150, 75, 25, 50, 100, 125, 175, 200, 225, 250], we want to find the maximum sum of elements that does not exceed K. We start by sorting the array in descending order. Then, we initialize two pointers, one at the start and one at the end of the array. We move the end pointer towards the start pointer until the sum of elements from the start pointer to the end pointer exceeds K. The maximum sum of elements that does not exceed K is the sum of elements from the start pointer to the second last pointer.
Input
[75, 25, 50, 100, 125, 175, 200, 225, 250]
Output
75
Explanation: Step-by-step: Given the input array [75, 25, 50, 100, 125, 175, 200, 225, 250], we want to find the maximum sum of elements that does not exceed K. We start by sorting the array in descending order. Then, we initialize two pointers, one at the start and one at the end of the array. We move the end pointer towards the start pointer until the sum of elements from the start pointer to the end pointer exceeds K. The maximum sum of elements that does not exceed K is the sum of elements from the start pointer to the second last pointer.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Recursively build the subset, aborting any branch where the partial sum exceeds the target, and optionally memoize (index, remaining) states to avoid recomputation.
Brute Force Approach
Generate every possible subset of the sequence and check if any subset sums to the target value.
Verified Code Solutions
function solution(nums, K) {
nums.sort((a, b) => b - a);
let left = 0, right = nums.length - 1, maxSum = 0;
while (left <= right) {
let sum = 0;
while (left <= right && sum + nums[left] <= K) {
sum += nums[left];
left++;
}
maxSum = Math.max(maxSum, sum);
right--;
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
sort(nums.begin(), nums.end(), greater<int>());
int left = 0, right = nums.size() - 1, maxSum = 0;
while (left <= right) {
int sum = 0;
while (left <= right && sum + nums[left] <= K) {
sum += nums[left];
left++;
}
maxSum = max(maxSum, sum);
right--;
}
return maxSum;
}
};class Solution {
public int solution(int[] nums, int K) {
Arrays.sort(nums);
int left = 0, right = nums.length - 1, maxSum = 0;
while (left <= right) {
int sum = 0;
while (left <= right && sum + nums[left] <= K) {
sum += nums[left];
left++;
}
maxSum = Math.max(maxSum, sum);
right--;
}
return maxSum;
}
}def solution(nums, K):
nums.sort(reverse=True)
left, right, maxSum = 0, len(nums) - 1, 0
while left <= right:
sum = 0
while left <= right and sum + nums[left] <= K:
sum += nums[left]
left += 1
maxSum = max(maxSum, sum)
right -= 1
return maxSumfunction solution(nums, K) {
nums.sort((a, b) => b - a);
let left = 0, right = nums.length - 1, maxSum = 0;
while (left <= right) {
let sum = 0;
while (left <= right && sum + nums[left] <= K) {
sum += nums[left];
left++;
}
maxSum = Math.max(maxSum, sum);
right--;
}
return maxSum;
}Asked in Top Tech Interviews
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