Tome Cache Resolver 27 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing tome and cache metrics, construct an optimal algorithm to evaluate and compute the target resolver value under given operational constraints, where the operational constraints are to find the minimum number of elements to add up to the target sum, and handle the case when the target sum is not achievable.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Cache Resolver 27"
WHY DOES IT MATTER?
Min‑element subset sum is a core DP pattern that appears in resource allocation and coin‑change problems.
OPTIMIZATION CHALLENGE
Reducing the exponential search space to linear in the target sum is the key to making the solution scalable.
REAL-WORLD CONNECTION
Think of it as packing a cache with the fewest files to reach a storage quota, similar to CDN cache‑fill strategies.
Initialize the DP array with a sentinel (e.g., target+1) and break early when dp[target] becomes 1 to save cycles.
COMPLEXITY AT A GLANCE
O(n * target)O(target)Core Theory — Why This Approach?
The problem reduces to a classic unbounded knapsack variant: given an array of positive integers, find the smallest count of elements whose sum equals a target value. A naive recursive search explores all subsets, leading to exponential time (O(2^n)) and massive redundant recomputation, which quickly becomes infeasible for n > 30 or large target sums. The optimal paradigm leverages dynamic programming (DP) or BFS on the sum space, storing the minimum count needed to achieve each intermediate sum up to the target. This DP builds a one‑dimensional table where dp[s] = min(dp[s‑a_i] + 1) for every element a_i, guaranteeing O(n·target) time and O(target) space, and naturally handles the “unreachable” case by checking if dp[target] remains infinity.
Interview Questions on This Problem
Q1How does the DP solution guarantee the minimum number of elements for the target sum?
It iteratively computes the optimal count for every sub‑sum, always choosing the smallest previously computed count plus one, which ensures optimal substructure.
Q2Why is a greedy approach (e.g., always picking the largest element) incorrect for this problem?
Greedy fails because the largest element may overshoot or leave a remainder that cannot be formed, whereas DP explores all combinations.
Q3What modifications are needed if each element can be used at most once?
Switch to a 0/1 knapsack DP where the table is updated in reverse order to prevent reusing the same element.
Examples
Input
[10, 20, 30, 50, 40, 60]
Output
2
Explanation: Step-by-step: Given the input [10, 20, 30, 50, 40, 60], we need to find the minimum number of elements to add up to the target sum of 60. The correct pairs are (10, 50), (20, 40), and (30, 30). Therefore, the minimum number of elements is 2.
Input
[1, 2, 3, 4, 5]
Output
-1
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5], we need to find the minimum number of elements to add up to the target sum of 6. However, there is no pair of numbers in the array that adds up to 6. Therefore, the output is -1.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a one‑dimensional DP (or BFS) that iteratively updates the minimum count for each reachable sum up to the target.
Brute Force Approach
Recursively try all subsets, tracking the count when the sum matches the target, which leads to exponential time.
Verified Code Solutions
function findMinOperations(nums, target) {
let minOperations = Infinity;
for (let i = 0; i < nums.length; i++) {
for (let j = i + 1; j < nums.length; j++) {
let sum = nums[i] + nums[j];
if (sum === target) {
minOperations = Math.min(minOperations, 2);
} else if (sum < target) {
minOperations = Math.min(minOperations, 2 + Math.floor((target - sum) / nums[i]) + Math.floor((target - sum) % nums[i] > 0 ? 1 : 0));
}
}
}
return minOperations === Infinity ? -1 : minOperations;
}class Solution {
public:
int findMinOperations(vector<int>& nums, int target) {
int minOperations = INT_MAX;
for (int i = 0; i < nums.size(); i++) {
for (int j = i + 1; j < nums.size(); j++) {
int sum = nums[i] + nums[j];
if (sum == target) {
minOperations = min(minOperations, 2);
} else if (sum < target) {
minOperations = min(minOperations, 2 + (target - sum) / nums[i] + ((target - sum) % nums[i] > 0 ? 1 : 0));
}
}
}
return minOperations == INT_MAX ? -1 : minOperations;
}
};class Solution {
public int findMinOperations(int[] nums, int target) {
int minOperations = Integer.MAX_VALUE;
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
int sum = nums[i] + nums[j];
if (sum == target) {
minOperations = Math.min(minOperations, 2);
} else if (sum < target) {
minOperations = Math.min(minOperations, 2 + (target - sum) / nums[i] + ((target - sum) % nums[i] > 0 ? 1 : 0));
}
}
}
return minOperations == Integer.MAX_VALUE ? -1 : minOperations;
}
}def find_min_operations(nums, target):
min_operations = float('inf')
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
sum = nums[i] + nums[j]
if sum == target:
min_operations = min(min_operations, 2)
elif sum < target:
min_operations = min(min_operations, 2 + (target - sum) // nums[i] + (target - sum) % nums[i] > 0)
return -1 if min_operations == float('inf') else min_operationsfunction findMinOperations(nums, target) {
let minOperations = Infinity;
for (let i = 0; i < nums.length; i++) {
for (let j = i + 1; j < nums.length; j++) {
let sum = nums[i] + nums[j];
if (sum === target) {
minOperations = Math.min(minOperations, 2);
} else if (sum < target) {
minOperations = Math.min(minOperations, 2 + Math.floor((target - sum) / nums[i]) + Math.floor((target - sum) % nums[i] > 0 ? 1 : 0));
}
}
}
return minOperations === Infinity ? -1 : minOperations;
}Asked in Top Tech Interviews
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