Monotonic Threshold Span Resolver 2 — Problem Statement & Solution Guide

Dynamic ProgrammingHardDigit DP

TimeO(N * 2 * 2)

SpaceO(2 * 2)

Problem Description

Given a high-dimensional input dataset or state graph of length N, calculate the optimal result using the Digit DP algorithm. The output should be the sum of all elements in the input dataset, but using a 3D DP table.

DSA Pattern Breakdown

"Monotonic Threshold Span Resolver 2"

hard

WHY DOES IT MATTER?

Digit DP abstracts complex numeric constraints into a handful of boolean flags, turning an exponential search into a linear scan. This pattern is indispensable when the problem asks for counts or sums over numbers that must obey digit‑wise rules, such as monotonic thresholds, digit sums, or range limits.

OPTIMIZATION CHALLENGE

The key insight is that only three bits of information are needed to describe the entire history: position, threshold compliance, and span activity. By collapsing all other details, we shrink the state space from exponential to O(N * 2 * 2), enabling a 3‑D DP that runs in linear time.

REAL-WORLD CONNECTION

Think of a distributed rate‑limiter that must enforce a sliding window of requests per user. The DP flags correspond to whether the current request is within the allowed window (active) and whether the quota has been exceeded (tight). The DP table aggregates the total allowed requests much like the limiter aggregates traffic across time buckets.

When coding the DP, pre‑allocate a static int64[2][2] array and toggle between two layers. Use clear variable names like 'tight' and 'inSpan' to avoid confusion, and always apply modulo operations (if required) after each addition to keep numbers within 64‑bit limits.

COMPLEXITY AT A GLANCE

⏱ Time:O(N * 2 * 2)

💾 Space:O(2 * 2)

Core Theory — Why This Approach?

Digit DP (also known as DP on the representation of numbers) transforms a seemingly combinatorial counting problem into a state‑space traversal over the digits of the input. By encoding constraints such as monotonic thresholds or span limits into a small set of flags, we can reuse sub‑problem results across overlapping prefixes, turning exponential brute‑force into polynomial time. In the "Monotonic Threshold Span Resolver 2" problem the state is three‑dimensional: the current position in the high‑dimensional dataset, a flag indicating whether we have already violated the monotonic threshold, and a flag tracking if the current span is active. This compact representation captures all necessary history without storing the entire prefix, which is why naive enumeration of all 2^N subsets fails for large N.

The optimal paradigm builds a DP table dp[pos][tight][active] where 'pos' iterates over the dataset indices, 'tight' records if the prefix so far respects the monotonic threshold, and 'active' records whether we are inside a valid span. Transitioning from one position to the next involves either extending the current span (if allowed) or closing it and possibly starting a new one. Because each transition depends only on the three flags, the total number of states is O(N * 2 * 2), and each state processes a constant number of digit choices, yielding linear‑time performance. This reduction from exponential to linear is the hallmark of Digit DP and is essential for handling inputs up to 10^5 or higher.

Interview Questions on This Problem

Q1How does Digit DP differ from classic DP, and why is it suitable for problems with monotonic thresholds?

Digit DP operates on the digit (or element) representation of a number or sequence, encoding constraints as state flags that capture prefix properties. Unlike classic DP which often works on sub‑arrays or sub‑problems directly, Digit DP leverages the limited range of each digit (e.g., 0‑9) to bound the state space. For monotonic thresholds, the 'tight' flag records whether the current prefix already exceeds the threshold, allowing us to prune invalid continuations early and keep the DP size small.

Q2Explain how you would compress a 3‑dimensional DP table to O(N) space for this problem.

Since the transition only depends on the previous position, we can roll the first dimension: maintain two 2‑D arrays dpPrev[tight][active] and dpCurr[tight][active]. After processing position i, we assign dpPrev = dpCurr and reuse the same memory for the next iteration, reducing space from O(N*4) to O(4) = O(1) per position, i.e., O(N) overall when accounting for the loop.

Q3A candidate suggests using recursion with memoization for the DP. What pitfalls should you watch for in a production‑grade solution?

Recursive memoization can cause stack overflow for N up to 10^5 and may incur significant overhead due to function calls and hashmap lookups. Iterative bottom‑up DP avoids recursion depth limits and typically runs faster because the state space is tiny and can be stored in plain arrays, which also simplifies memory management and cache locality.

Examples

Example 1

Input

[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Output

Explanation: Step-by-step: Given a 3D array of size 3x3x3, we initialize a 3D DP table dp of size 3x3x3 with all elements as 0. We then iterate over each element in the input array, and for each element, we update the corresponding dp table entry with the sum of the current element and the maximum of the two elements directly above it in the dp table. Finally, we return the sum of all elements in the dp table, which represents the optimal result.

Example 2

Input

[[10, 20, 30], [40, 50, 60], [70, 80, 90]]

Output

Constraints

1 <= N <= 2 * 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity: O(N log N) or O(N log^2 N)
Space Complexity: O(N)

Optimal Approach & Strategy

Use a 3‑dimensional Digit DP table dp[pos][tight][active] to propagate sums in linear time, rolling the first dimension to achieve O(1) extra space.

Brute Force Approach

Enumerate every possible subset of indices, check the monotonic threshold for each, and sum the elements of valid subsets.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N * 2 * 2)

function solution(nums) {
   const n = nums.length;
   const m = nums[0].length;
   const p = nums[0][0].length;
   const dp = new Array(n).fill(0).map(() => new Array(m).fill(0).map(() => new Array(p).fill(0)));
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < m; j++) {
           for (let k = 0; k < p; k++) {
               if (i === 0 && j === 0 && k === 0) {
                   dp[i][j][k] = nums[i][j][k];
               } else {
                   dp[i][j][k] = nums[i][j][k] + Math.max(dp[i-1][j][k], dp[i][j-1][k], dp[i][j][k-1]);
               }
           }
       }
   }
   let result = 0;
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < m; j++) {
           for (let k = 0; k < p; k++) {
               result += dp[i][j][k];
           }
       }
   }
   return result;
}

class Solution {
   public:
       int solution(int*** nums) {
           int n = nums->length;
           int m = nums->length;
           int p = nums->length;
           int*** dp = new int**[n];
           for (int i = 0; i < n; i++) {
               dp[i] = new int*[m];
               for (int j = 0; j < m; j++) {
                   dp[i][j] = new int[p];
               }
           }
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < m; j++) {
                   for (int k = 0; k < p; k++) {
                       if (i == 0 && j == 0 && k == 0) {
                           dp[i][j][k] = nums[i][j][k];
                       } else {
                           dp[i][j][k] = nums[i][j][k] + std::max(dp[i-1][j][k], dp[i][j-1][k], dp[i][j][k-1]);
                       }
                   }
               }
           }
           int result = 0;
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < m; j++) {
                   for (int k = 0; k < p; k++) {
                       result += dp[i][j][k];
                   }
               }
           }
           return result;
       }
   };

class Solution {
   public int solution(int[][][] nums) {
       int n = nums.length;
       int m = nums[0].length;
       int p = nums[0][0].length;
       int[][][] dp = new int[n][m][p];
       for (int i = 0; i < n; i++) {
           for (int j = 0; j < m; j++) {
               for (int k = 0; k < p; k++) {
                   if (i == 0 && j == 0 && k == 0) {
                       dp[i][j][k] = nums[i][j][k];
                   } else {
                       dp[i][j][k] = nums[i][j][k] + Math.max(dp[i-1][j][k], dp[i][j-1][k], dp[i][j][k-1]);
                   }
               }
           }
       }
       int result = 0;
       for (int i = 0; i < n; i++) {
           for (int j = 0; j < m; j++) {
               for (int k = 0; k < p; k++) {
                   result += dp[i][j][k];
               }
           }
       }
       return result;
   }
}

def solution(nums):
   n = len(nums)
   m = len(nums[0])
   p = len(nums[0][0])
   dp = [[[0 for _ in range(p)] for _ in range(m)] for _ in range(n)]
   for i in range(n):
       for j in range(m):
           for k in range(p):
               if i == 0 and j == 0 and k == 0:
                   dp[i][j][k] = nums[i][j][k]
               else:
                   dp[i][j][k] = nums[i][j][k] + max(dp[i-1][j][k], dp[i][j-1][k], dp[i][j][k-1])
   result = 0
   for i in range(n):
       for j in range(m):
           for k in range(p):
               result += dp[i][j][k]
   return result

function solution(nums) {
   const n = nums.length;
   const m = nums[0].length;
   const p = nums[0][0].length;
   const dp = new Array(n).fill(0).map(() => new Array(m).fill(0).map(() => new Array(p).fill(0)));
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < m; j++) {
           for (let k = 0; k < p; k++) {
               if (i === 0 && j === 0 && k === 0) {
                   dp[i][j][k] = nums[i][j][k];
               } else {
                   dp[i][j][k] = nums[i][j][k] + Math.max(dp[i-1][j][k], dp[i][j-1][k], dp[i][j][k-1]);
               }
           }
       }
   }
   let result = 0;
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < m; j++) {
           for (let k = 0; k < p; k++) {
               result += dp[i][j][k];
           }
       }
   }
   return result;
}

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Monotonic Threshold Span Resolver 2 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Dynamic Programming

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Monotonic Threshold Span Resolver 2 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor