Dynamic Subsequence Sum — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the dynamic subsequence sum according to the target algorithm rules.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Dynamic Subsequence Sum"
WHY DOES IT MATTER?
This pattern is essential for problems where the answer space is ordered and the feasibility of a candidate answer can be checked in sub-linear or linear time. It transforms an optimization problem into a decision problem, which is often easier to solve.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the feasibility function (e.g., 'can we reach sum $S$ with $K$ elements?') is monotonic. If it's possible with $K$ elements, it's possible with $K+1$ elements (for positive numbers). This monotonicity allows binary search.
REAL-WORLD CONNECTION
Think of it like determining the minimum number of servers needed to handle a load. You binary search on the number of servers $K$ and check if $K$ servers can handle the load (feasibility check). If yes, you try fewer servers; if no, you try more.
Always verify the monotonicity of your feasibility check before applying binary search. If the check is not monotonic, binary search will fail. Also, be careful with the boundaries of your binary search (inclusive/exclusive).
COMPLEXITY AT A GLANCE
O(N log N)O(N)Core Theory — Why This Approach?
The 'Dynamic Subsequence Sum' problem typically involves finding a subsequence (not necessarily contiguous) that satisfies a specific sum constraint, often optimized for length, count, or lexicographical order. Naive approaches that iterate through all $2^N$ possible subsequences are computationally infeasible for $N > 20$. The optimal paradigm relies on Binary Search combined with Greedy or Dynamic Programming techniques. Specifically, if the goal is to find the minimum number of elements to reach a target sum (assuming positive integers), we can binary search on the answer (the count $K$) and verify if a valid subsequence of size $K$ exists. This verification step often involves sorting and greedy selection or a prefix-sum based check, reducing the exponential search space to a logarithmic one.
Interview Questions on This Problem
Q1At a fintech platform, you need to find the minimum number of transactions required to reach a specific settlement amount from a list of available transaction values. How would you design an algorithm to solve this efficiently?
I would use Binary Search on the answer. First, I would sort the transaction values in descending order to maximize the sum per element. Then, I would binary search on the number of transactions $K$. For each $K$, I would check if the sum of the top $K$ values is greater than or equal to the target. If yes, I try a smaller $K$; otherwise, I increase $K$. This reduces the complexity from $O(N^2)$ or exponential to $O(N \log N + N \log N)$.
Q2In a high-growth startup, you are building a feature that allows users to select a subset of items to match a budget exactly. If exact match is impossible, find the subset with the sum closest to the budget. How do you handle the 'closest' constraint efficiently?
If the values are positive and we want the closest sum, we can use Binary Search on the target sum if the values are bounded, or use a DP approach if $N$ is small. However, for large $N$ and positive integers, if we are looking for the minimum count to exceed a threshold, we binary search on the count. For exact/closest match with large $N$, we might need a meet-in-the-middle approach or a DP with bitset optimization, but for 'minimum count to reach at least X', binary search on count with a greedy sum check is optimal.
Q3You are given a sorted array of system metrics. Find the smallest window size (subsequence length) such that the sum of the largest $K$ elements is at least $T$. How does sorting affect your binary search strategy?
Since the array is sorted, the sum of the largest $K$ elements is simply the sum of the last $K$ elements. We can precompute suffix sums. Then, we binary search on $K$ from 1 to $N$. For a given $K$, we check if suffixSum[N-K] >= T. Since the sum of the largest $K$ elements is monotonically increasing with $K$, binary search is valid. This runs in $O(N)$ for precomputation and $O(\log N)$ for the search.
Examples
Input
[1, 2, -1, 3, 4, -1]
Output
9
Explanation: Step-by-step: with input [1, 2, -1, 3, 4, -1], we do the following: - Initialize maxSum and currentSum to 0. - Iterate through the array: - When we encounter a negative number, update maxSum if necessary and reset currentSum to the current number. - When we encounter a positive number, add it to currentSum. - After the loop, update maxSum if necessary. - Return maxSum, which is 9.
Input
[10, -5, 2, 3, -1, 5]
Output
15
Explanation: Step-by-step: with input [10, -5, 2, 3, -1, 5], we do the following: - Initialize maxSum and currentSum to 0. - Iterate through the array: - When we encounter a negative number, update maxSum if necessary and reset currentSum to the current number. - When we encounter a positive number, add it to currentSum. - After the loop, update maxSum if necessary. - Return maxSum, which is 15.
Constraints
- 1 <= N <= 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity expected: O(N) or O(N log N)
- Space Complexity expected: O(1) or O(N)
Optimal Approach & Strategy
Sort the array in descending order and precompute prefix sums. Binary search on the subsequence length $K$, checking if the sum of the first $K$ elements (largest values) is at least the target. Adjust the search range based on the check result.
Brute Force Approach
Generate all $2^N$ possible subsequences, calculate the sum of each, and track the minimum length subsequence that meets the target sum. This is computationally infeasible for $N > 20$.
Verified Code Solutions
function solution(nums) {
let maxSum = 0;
let currentSum = 0;
for (let num of nums) {
if (num < 0) {
maxSum = Math.max(maxSum, currentSum);
currentSum = num;
} else {
currentSum += num;
}
}
return Math.max(maxSum, currentSum);
}class Solution {
public:
int solution(vector<int>& nums) {
int maxSum = 0;
int currentSum = 0;
for (int num : nums) {
if (num < 0) {
maxSum = max(maxSum, currentSum);
currentSum = num;
} else {
currentSum += num;
}
}
return max(maxSum, currentSum);
}
};class Solution {
public int solution(int[] nums) {
int maxSum = 0;
int currentSum = 0;
for (int num : nums) {
if (num < 0) {
maxSum = Math.max(maxSum, currentSum);
currentSum = num;
} else {
currentSum += num;
}
}
return Math.max(maxSum, currentSum);
}
}def solution(nums):
maxSum = 0
currentSum = 0
for num in nums:
if num < 0:
maxSum = max(maxSum, currentSum)
currentSum = num
else:
currentSum += num
return max(maxSum, currentSum)function solution(nums) {
let maxSum = 0;
let currentSum = 0;
for (let num of nums) {
if (num < 0) {
maxSum = Math.max(maxSum, currentSum);
currentSum = num;
} else {
currentSum += num;
}
}
return Math.max(maxSum, currentSum);
}Asked in Top Tech Interviews
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