Tome Signal Evaluator 10 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing tome and signal metrics, construct an optimal algorithm to evaluate and compute the target evaluator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Signal Evaluator 10"
WHY DOES IT MATTER?
Using prefix sums with a hash map turns an O(n^2) counting problem into O(n) time, essential for large datasets.
OPTIMIZATION CHALLENGE
The key is to avoid re‑scanning substrings by remembering the first occurrence of each cumulative difference.
REAL-WORLD CONNECTION
Similar to detecting equal traffic flow in network packets, where you track cumulative differences to spot balanced intervals.
Pre‑allocate the hash map size based on input length to reduce rehashing overhead in production code.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem reduces to finding the longest substring where the counts of two distinct characters are equal. A naive approach would enumerate all substrings and count characters, leading to O(n^2) time and O(1) space, which becomes infeasible for strings of length 10^5 or more. The optimal solution transforms the problem into a prefix sum difference array: assign +1 for one character and -1 for the other, then the difference between any two indices indicates the net balance. By storing the earliest occurrence of each prefix sum in a hash map, we can compute the maximum distance between equal sums in a single pass, achieving O(n) time and O(n) space.
This technique is a classic application of the “prefix sum + hash map” paradigm, often used for subarray sum problems and longest balanced substring problems. It avoids repeated counting by leveraging the fact that equal prefix sums imply equal numbers of the two characters between them. The hash map ensures constant‑time lookups for the earliest index of a given sum, turning a quadratic brute force into linear time.
Interview Questions on This Problem
Q1What is the time complexity of the optimal solution for finding the longest balanced substring of two characters?
O(n) time, because we traverse the string once and perform constant‑time hash map operations.
Q2Why does storing the earliest index of each prefix sum help in maximizing substring length?
The first occurrence gives the farthest possible right endpoint for that sum, ensuring the longest distance between equal sums.
Q3How would you modify the algorithm if the string contained more than two distinct characters?
You would need to track counts for all characters, which typically leads to a multi‑dimensional prefix or use hashing of count vectors, increasing complexity beyond linear.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50]
Output
939
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50], we first find the sum of all numbers in the array, which is 1225. Then, we find the sum of all numbers less than or equal to K, which is 6. Finally, we subtract the sum of numbers less than or equal to K from the sum of all numbers in the array to get the sum of numbers greater than K, which is 1219. However, this is incorrect. The correct approach is to find the sum of numbers greater than K by iterating over the array and adding numbers greater than K to a variable. The correct sum is 939 because 4+5+6+7+8+9 = 33, and 10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50 = 912, and 33 + 912 = 945, and 945 - (1+2+3) = 939.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]
Output
0
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25], we first find the sum of all numbers in the array, which is 300. Then, we find the sum of all numbers less than or equal to K, which is 15. Finally, we subtract the sum of numbers less than or equal to K from the sum of all numbers in the array to get the sum of numbers greater than K, which is 285. However, this is incorrect. The correct approach is to find the sum of numbers greater than K by iterating over the array and adding numbers greater than K to a variable. The correct sum is 0 because there are no numbers greater than 25 in the array.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Traverse once, maintain a running sum (+1 for one char, -1 for the other), and use a hash map to store the earliest index of each sum to compute maximum distance in O(n) time.
Brute Force Approach
Check every possible substring, count characters, and update the maximum length when counts match. This takes O(n^2) time.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (num > k) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int> nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
}def solution(nums, k):
sum = 0
for num in nums:
if num > k:
sum += num
return sumfunction solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (num > k) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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