Find the Count of Monotonic Pairs II - Practice Coding Problems

Find the Count of Monotonic Pairs II - Problem

Array Math Dynamic Programming Combinatorics Prefix Sum Hard

You are given an array of positive integers nums of length n.

We call a pair of non-negative integer arrays (arr1, arr2) monotonic if:

The lengths of both arrays are n.
arr1 is monotonically non-decreasing, in other words, arr1[0] <= arr1[1] <= ... <= arr1[n - 1].
arr2 is monotonically non-increasing, in other words, arr2[0] >= arr2[1] >= ... >= arr2[n - 1].
arr1[i] + arr2[i] == nums[i] for all 0 <= i <= n - 1.

Return the count of monotonic pairs. Since the answer may be very large, return it modulo 10⁹ + 7.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,3,2]

› Output: 4

💡 Note: Valid monotonic pairs: ([0,2,1],[2,1,1]), ([0,2,2],[2,1,0]), ([0,3,1],[2,0,1]), ([0,3,2],[2,0,0]). Each satisfies arr1 non-decreasing, arr2 non-increasing, and arr1[i] + arr2[i] = nums[i].

Example 2 — Single Element

$ Input: nums = [5]

› Output: 6

💡 Note: For single element, arr1[0] can be 0,1,2,3,4,5 and arr2[0] = 5-arr1[0]. All 6 combinations are valid since there are no monotonic constraints to check.

Example 3 — Identical Elements

$ Input: nums = [2,2]

› Output: 6

💡 Note: Valid pairs include ([0,0],[2,2]), ([0,1],[2,1]), ([0,2],[2,0]), ([1,1],[1,1]), ([1,2],[1,0]), ([2,2],[0,0]). Total count is 6.

Constraints

1 ≤ nums.length ≤ 2000
1 ≤ nums[i] ≤ 1000

Visualization

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Understanding the Visualization

Input Array

Given nums = [2,3,2]

Split Each Element

nums[i] = arr1[i] + arr2[i]

Apply Constraints

arr1 non-decreasing, arr2 non-increasing

Count Valid Pairs

Total monotonic pairs

Key Takeaway

🎯 Key Insight: Transform the splitting problem into dynamic programming with monotonic constraints

Asked in

G Google 15 f Facebook 12 a Amazon 8

The key insight is recognizing this as a constrained counting problem where we split each nums[i] into arr1[i] + arr2[i] with monotonic constraints. The optimal approach uses dynamic programming with mathematical optimization to achieve O(n × max(nums)) time complexity by tracking valid states efficiently.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Recursion	O(2^n)	O(n)	Try all possible values for arr1[i] and arr2[i] at each position
Optimized DP with Mathematical Insight	O(n × max(nums))	O(max(nums))	Use combinatorics to reduce the problem dimension and optimize transitions
Dynamic Programming	O(n × max(nums)²)	O(n × max(nums)²)	Use memoization to avoid recomputing the same subproblems

Brute Force Recursion — Algorithm Steps

For each position i, try all values 0 ≤ arr1[i] ≤ nums[i]
Set arr2[i] = nums[i] - arr1[i]
Check monotonic constraints and recurse
Count all valid complete assignments

Visualization

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Step-by-Step Walkthrough

Position 0

Try all arr1[0] values, compute arr2[0]

Position 1

For each valid previous choice, try next values

Count Valid

Sum all complete valid assignments

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

const int MOD = 1000000007;

long long backtrack(int* nums, int n, int pos, int prevArr1, long long prevArr2) {
    if (pos == n) return 1;
    
    long long count = 0;
    int start = (prevArr1 > 0) ? prevArr1 : 0;
    for (int arr1Val = start; arr1Val <= nums[pos]; arr1Val++) {
        int arr2Val = nums[pos] - arr1Val;
        if (arr2Val >= 0 && arr2Val <= prevArr2) {
            count = (count + backtrack(nums, n, pos + 1, arr1Val, arr2Val)) % MOD;
        }
    }
    return count;
}

int solution(int* nums, int n) {
    return backtrack(nums, n, 0, 0, LLONG_MAX);
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Parse JSON array manually
    int nums[1000];
    int n = 0;
    char* ptr = line + 1; // Skip '['
    while (*ptr && *ptr != ']') {
        if (*ptr >= '0' && *ptr <= '9') {
            nums[n++] = strtol(ptr, &ptr, 10);
        } else {
            ptr++;
        }
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

Each position has up to max(nums) choices, leading to exponential branching

⚠ Quadratic Growth

Space Complexity

O(n)

Recursion stack depth is n

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Recursion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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