2143. Choose Numbers From Two Arrays in Range

Problem Description

The problem provides two arrays nums1 and nums2 of the same length n. The task is to find the number of different balanced ranges within these arrays. A balanced range l, r means that for every index i within this range, you can choose either nums1[i] or nums2[i] such that the sum of the selected elements from nums1 is equal to the sum of the selected elements from nums2.

Balanced ranges [l1, r1] and [l2, r2] are considered different if either the starting points l1 and l2 differ, the ending points r1 and r2 differ, or there is at least one index i where nums1[i] is chosen in the first range, and nums2[i] is chosen in the second range, or the other way around.

The solution must return the count of these distinctive balanced ranges modulo 10^9 + 7.

Intuition

The solution is based on dynamic programming. The primary intuition is to track the difference between sums of selected elements from nums1 and nums2 at each index i when considering all possible subranges ending at that index. Essentially, you want to know how many ways you can achieve a particular sum difference up to a certain point, which in turn will help you decide how many balanced subranges you can form.

Here's how you arrive at the solution:

Create a list of lists f, where f[i][j] stores the number of ways to achieve a sum difference of j - s2 using subranges that end at index i. s2 is the total sum of nums2, and this acts as a balance point to avoid negative indices in f due to potential negative differences.
Iterate through each index i of the arrays, and for each i, consider adding the current value a from nums1 or subtracting the current value b from nums2 to all previously computed sum differences to update the number of ways to achieve new differences at i based on those at index i - 1. That is, f[i][j] is updated by adding f[i - 1][j - a] and f[i - 1][j + b] considering the bounds where these indices do not go beyond the size of f.
Each time, keep track of the number of balanced subranges, which essentially corresponds to the value of f[i][s2] because a sum difference of 0 indicates that the sums from both nums1 and nums2 are equal.
The final answer is the accumulated count of balanced subranges modulo 10^9 + 7 to handle the large numbers as mentioned in the problem.

This approach efficiently uses the concept of prefix sums and dynamic programming to solve the problem in polynomial time.

Learn more about Dynamic Programming patterns.

Not Sure What to Study? Take the 2-min Quiz to Find Your Missing Piece：

Which technique can we use to find the middle of a linked list?

Fast-slow pointersTwo pointers moving in opposite directionBinary searchDepth first search

I don’t know

Solution Approach

The implementation uses dynamic programming to keep track of all the possible sum differences between nums1 and nums2 as you iterate over them. Here is a detailed walkthrough:

Initialization: Two sums, s1 and s2, are calculated to represent the total sums of nums1 and nums2, respectively. A 2D list f with dimensions [n][s1 + s2 + 1] is created. This list will store the number of ways to obtain a sum difference at various j points, for each i. The +1 accommodates zero difference.
Calculating Ways for Differences: As we iterate through nums1 and nums2 with index i, we increase f[i][a + s2] and f[i][-b + s2] by 1. This reflects the fact that selecting a from nums1 or b from nums2 at the current index contributes to one way of making the sum difference of a - b (indexed from -b + s2 to a + s2 to shift the negative range).
Updating Counts: If i is greater than 0, we have previous states to consider. The dynamic programming aspect comes in:
- We update f[i][j] by adding the number of ways to achieve a difference of j-a from the previous index i-1 to the number of ways to obtain j at i, ensuring that j is large enough (j >= a).
- We also add the number of ways to achieve a difference of j+b from index i-1 to the number of ways to obtain j at i, making sure we don't exceed the range of sum differences (j+b < s1 + s2 + 1).
Here, j is the index representing the possible sum differences, a is the element from nums1, and b is the element from nums2.
Counting Balanced Ranges: The f[i][s2] entry contains the number of ways to have a zero sum difference up to index i, which corresponds to a balanced range. We add f[i][s2] to ans, the accumulated total of such balanced ranges, and apply % mod to ensure the result stays within the required modulo.
Return Result: The variable ans stores the final count and is returned to represent the number of different balanced ranges (modulo 10^9 + 7).

This approach uses a dynamic table f and iterates through each element of nums1 and nums2 once, updating the counts of sum differences as it goes. The table f stores intermediate results that are re-used, which is a classic feature of dynamic programming, and it uses the modulo operator to manage large numbers efficiently.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

Which of the following is the prefix sum of array [1, 2, 3, 4, 5]?

[1, 2, 3, 4, 5][1, 3, 6, 10, 15][15, 14, 12, 9, 5][5, 4, 3, 2, 1]

I don’t know

Example Walkthrough

Let's consider the arrays nums1 = [1,2,3] and nums2 = [2,1,2], and walk through the described solution approach.

Initial Setup: First, we calculate the total sums.

s1 = 1 + 2 + 3 = 6
s2 = 2 + 1 + 2 = 5

We initialize the 2D list f with dimensions [3][6 + 5 + 1] or [3][12], as we have 3 elements and the sum difference can range from -5 to 6. We index f from 0, so the actual index for a zero sum difference is 5, which is s2.

Calculating Ways for Differences:

At index 0, with nums1[0] = 1 and nums2[0] = 2:
- f[0][1 + 5] or f[0][6] represents choosing 1 from nums1 and increases by 1.
- f[0][-2 + 5] or f[0][3] represents choosing 2 from nums2 and increases by 1.
At index 1, with nums1[1] = 2 and nums2[1] = 1:
- f[1][2 + 5] or f[1][7] represents choosing 2 from nums1. Since i > 0, we add the values from f[0][7-2] (or f[0][5], which is currently 0) and f[0][7+1] (or f[0][8], which does not exist and is considered 0). Hence, f[1][7] increases by 1.
- f[1][-1 + 5] or f[1][4] represents choosing 1 from nums2 and we perform the same additions as above, so f[1][4] increases by 1.
At index 2, with nums1[2] = 3 and nums2[2] = 2:
- We update f[2][3 + 5] or f[2][8] and f[2][-2 + 5] or f[2][3] similarly.

Updating Counts:

For every j from 0 to 11 (which corresponds to the range of -5 to 6), update f[i][j] by adding the values from the last index i - 1 with the differences of j - nums1[i] and j + nums2[i], if those indices are valid.

Counting Balanced Ranges:

For each index i, we add f[i][5] to ans, because f[i][5] represents the count of balanced ranges up to index i. For instance, f[2][5] will tell us the number of ways we can have a balanced subrange ending at index 2.

Return Result:

After these steps, ans, now containing the total count of balanced ranges, is returned as the answer modulo 10^9 + 7.

Plugging in the numbers:

After index 0, f[0][6] = 1 and f[0][3] = 1. No balanced range yet.
After index 1, f[1][7] = 1 and f[1][4] = 1. No balanced range yet.
After index 2, f[2][8] and f[2][3] are updated. We find f[2][5] = 1 indicating one balanced range [0,2].

Our ans would be 1 modulo 10^9 + 7, which is just 1, since we found one balanced range. This would be the final returned value.

Solution Implementation

1from typing import List
2
3class Solution:
4    def countSubranges(self, nums1: List[int], nums2: List[int]) -> int:
5        length = len(nums1)  # store the length of nums1 and nums2, which should be the same
6        sum1, sum2 = sum(nums1), sum(nums2)  # calculate the sum of the elements in nums1 and nums2
7        # create a 2D list to keep track of counts while avoiding index-out-of-range errors
8        counts = [[0] * (sum1 + sum2 + 1) for _ in range(length)]
9        total_count = 0  # initialize the result to accumulate the total count of valid subranges
10        modulo = 10**9 + 7  # the value for modulo operation to avoid large integers
11
12        # iterate through both lists in parallel using enumerate to get both index and elements
13        for i, (num1, num2) in enumerate(zip(nums1, nums2)):
14            counts[i][num1 + sum2] += 1  # increment the count where the first element is picked from nums1
15            counts[i][-num2 + sum2] += 1  # increment the count where the first element is picked from nums2
16
17            # if not on the first index, update the counts array based on previous counts
18            if i:
19                for j in range(sum1 + sum2 + 1):
20                    if j >= num1:
21                        counts[i][j] = (counts[i][j] + counts[i - 1][j - num1]) % modulo
22                    if j + num2 < sum1 + sum2 + 1:
23                        counts[i][j] = (counts[i][j] + counts[i - 1][j + num2]) % modulo
24
25            # update the total count for subranges that sum up to zero difference
26            total_count = (total_count + counts[i][sum2]) % modulo
27
28        return total_count  # return the total count of valid subranges
29
30# The function countSubranges calculates the number of subranges where the sum of selected elements from nums1
31# equals the sum of selected elements from nums2. It modifies the subproblem's state space to make it 
32# solvable using dynamic programming, ensuring that each choice at every step is either to include 
33# an element from nums1 or an element from nums2.
34

1class Solution {
2    public int countSubranges(int[] nums1, int[] nums2) {
3        int n = nums1.length; // Get the length of the input arrays.
4        int sumNums1 = Arrays.stream(nums1).sum(); // Sum of all elements in nums1.
5        int sumNums2 = Arrays.stream(nums2).sum(); // Sum of all elements in nums2.
6  
7        // Create a 2D array to store the number of ways to form subranges.
8        int[][] dp = new int[n][sumNums1 + sumNums2 + 1];
9        int answer = 0; // Initialize the answer variable to store the total count of subranges.
10        final int MOD = (int) 1e9 + 7; // Define the modulo value.
11  
12        // Iterate through each element in both arrays.
13        for (int i = 0; i < n; ++i) {
14            int num1 = nums1[i], num2 = nums2[i]; // Get the current elements from both arrays.
15            dp[i][num1 + sumNums2]++; // Increment the number of ways to achieve this sum with only the current element from nums1.
16            dp[i][-num2 + sumNums2]++; // Increment the number of ways to achieve this sum with only the current element from nums2.
17          
18            // If not in the first iteration, update the dp array based on previous subranges.
19            if (i > 0) {
20                for (int j = 0; j <= sumNums1 + sumNums2; ++j) {
21                    if (j >= num1) {
22                        dp[i][j] = (dp[i][j] + dp[i - 1][j - num1]) % MOD; // Add ways to achieve this sum including the current number from nums1.
23                    }
24                    if (j + num2 <= sumNums1 + sumNums2) {
25                        dp[i][j] = (dp[i][j] + dp[i - 1][j + num2]) % MOD; // Add ways to achieve this sum including the current number from nums2.
26                    }
27                }
28            }
29            answer = (answer + dp[i][sumNums2]) % MOD; // Update the answer with the number of ways to achieve zero sum difference.
30        }
31        return answer; // Return the total count of subranges with zero sum difference.
32    }
33}
34

1#include <vector>
2#include <numeric>
3#include <cstring>
4
5class Solution {
6public:
7    int countSubranges(vector<int>& nums1, vector<int>& nums2) {
8        int n = nums1.size(); // size of the input arrays
9        int sum1 = accumulate(nums1.begin(), nums1.end(), 0); // sum of nums1
10        int sum2 = accumulate(nums2.begin(), nums2.end(), 0); // sum of nums2
11
12        // We'll use a dynamic programming array 'dp' to store the number of ways
13        // to get a sum taking first (i+1) elements where the sum is offset by sum2
14        // to handle negative sums.
15        int dp[n][sum1 + sum2 + 1];
16        memset(dp, 0, sizeof(dp)); // initialize dp array to 0
17
18        int ans = 0; // this will hold the final answer
19        const int mod = 1e9 + 7; // modulo value for the answer
20
21        // Calculate the number of subranges for each element
22        for (int i = 0; i < n; ++i) {
23            int a = nums1[i], b = nums2[i]; // Current elements from both arrays
24            dp[i][a + sum2]++; // If we take nums1[i], add to count
25            dp[i][-b + sum2]++; // If we take nums2[i], add to count
26
27            // Update the 'dp' array for the rest of the possible sums
28            if (i > 0) { // we skip the first element because there's nothing to accumulate from
29                for (int j = 0; j <= sum1 + sum2; ++j) {
30                    if (j >= a) {
31                        // Include the current nums1[i] in the subrange and add the count
32                        // from the previous subrange sum without current nums1[i]
33                        dp[i][j] = (dp[i][j] + dp[i - 1][j - a]) % mod;
34                    }
35                    if (j + b <= sum1 + sum2) {
36                        // Include the current nums2[i] in the subrange and add the count
37                        // from the previous subrange sum without current nums2[i]
38                        dp[i][j] = (dp[i][j] + dp[i - 1][j + b]) % mod;
39                    }
40                }
41            }
42
43            // Sum up the ways to achieve sum2 (offset sum is 0) for the current element
44            ans = (ans + dp[i][sum2]) % mod;
45        }
46
47        return ans; // Return the total number of valid subranges
48    }
49};
50

1function countSubranges(nums1: number[], nums2: number[]): number {
2    const lengthOfNums = nums1.length;
3    const sumOfNums1 = nums1.reduce((total, current) => total + current, 0);
4    const sumOfNums2 = nums2.reduce((total, current) => total + current, 0);
5    // Initialize dynamic programming table to store intermediate results
6    const dpTable: number[][] = Array(lengthOfNums)
7        .fill(0)
8        .map(() => Array(sumOfNums1 + sumOfNums2 + 1).fill(0));
9    const modulo = 1e9 + 7; // The modulo value to ensure results within integer limits
10    let countOfSubranges = 0; // Variable to keep the final count of subranges
11
12    // Iterate over each pair of elements from nums1 and nums2
13    for (let i = 0; i < lengthOfNums; ++i) {
14        const valueFromNums1 = nums1[i];
15        const valueFromNums2 = nums2[i];
16        // Increase the count for the subrange that only includes current element
17        dpTable[i][valueFromNums1 + sumOfNums2]++;
18        dpTable[i][-valueFromNums2 + sumOfNums2]++;
19
20        // If current index is not the first, calculate the count of subranges that end at the current index
21        if (i > 0) {
22            for (let j = 0; j <= sumOfNums1 + sumOfNums2; ++j) {
23                // If subrange can be extended by adding value from nums1
24                if (j >= valueFromNums1) {
25                    dpTable[i][j] = (dpTable[i][j] + dpTable[i - 1][j - valueFromNums1]) % modulo;
26                }
27                // If subrange can be extended by subtracting value from nums2
28                if (j + valueFromNums2 <= sumOfNums1 + sumOfNums2) {
29                    dpTable[i][j] = (dpTable[i][j] + dpTable[i - 1][j + valueFromNums2]) % modulo;
30                }
31            }
32        }
33        // Add the count of balanced subranges (where sum of nums1 elements equals sum of nums2 elements) to the answer
34        countOfSubranges = (countOfSubranges + dpTable[i][sumOfNums2]) % modulo;
35    }
36
37    // Return the total count of subranges found
38    return countOfSubranges;
39}
40

Not Sure What to Study? Take the 2-min Quiz：

What are the most two important steps in writing a depth first search function? (Select 2)

Determine state to pass down to recursive function callDetermine return valueDetermine how many recursive calls will be madeTracing the recursive calls

Time and Space Complexity

The given Python code aims to count the number of subranges in two arrays, nums1 and nums2, where for each subrange (i, j), the sum from i to j in nums1 is equal to the sum from i to j in nums2. The algorithm uses dynamic programming to keep track of the possible sums.

Time Complexity

To analyze the time complexity, we consider the number of operations performed:

The algorithm iterates over each pair (a, b) from nums1 and nums2.
Inside the outer loop that iterates over n, where n is the length of nums1 (or nums2), there is an inner loop that runs from 0 to s1 + s2, where s1 is the sum of all elements in nums1 and s2 is the sum of all elements in nums2. This means that the inner loop runs for O(s1 + s2) iterations for each i.
The inner loop operations consist of a constant number of arithmetic operations, each having a time complexity of O(1).

Combining this information, the total time complexity is O(n * (s1 + s2)), as there are n iterations in the outer loop and O(s1 + s2) operations for each iteration in the inner loop.

Space Complexity

For space complexity, we consider the storage used:

The algorithm allocates a 2D list f with n rows and s1 + s2 + 1 columns, where n is the length of the arrays and s1 + s2 is the sum of the elements in nums1 and nums2. Therefore, the space required for this list is O(n * (s1 + s2)).
Other variables used (n, s1, s2, ans, mod, a, b, i, j) require constant space, hence O(1).

This results in a total space complexity of O(n * (s1 + s2)), dominated by the space required for the 2D list f.

Learn more about how to find time and space complexity quickly using problem constraints.

Fast Track Your Learning with Our Quick Skills Quiz:

Given a sorted array of integers and an integer called target, find the element that equals to the target and return its index. Select the correct code that fills the ___ in the given code snippet.

1def binary_search(arr, target):
2    left, right = 0, len(arr) - 1
3    while left ___ right:
4        mid = (left + right) // 2
5        if arr[mid] == target:
6            return mid
7        if arr[mid] < target:
8            ___ = mid + 1
9        else:
10            ___ = mid - 1
11    return -1
12

1public static int binarySearch(int[] arr, int target) {
2    int left = 0;
3    int right = arr.length - 1;
4
5    while (left ___ right) {
6        int mid = left + (right - left) / 2;
7        if (arr[mid] == target) return mid;
8        if (arr[mid] < target) {
9            ___ = mid + 1;
10        } else {
11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16

1function binarySearch(arr, target) {
2    let left = 0;
3    let right = arr.length - 1;
4
5    while (left ___ right) {
6        let mid = left + Math.trunc((right - left) / 2);
7        if (arr[mid] == target) return mid;
8        if (arr[mid] < target) {
9            ___ = mid + 1;
10        } else {
11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16

<=, left, right<, left, right<=, right, left<, right, left

I don’t know

Overview

Basic Data Structures

Basic Algorithms

Overview

Sorted Array

Implicitly Sorted Array

Advanced

Speedrun

Introduction

DFS on Tree

Binary Search Tree

Advanced

Combinatorial Search

Pruning

Additional States

Aggregation and Memoization

Dedup

Additional Practices

Speedrun

Introduction

BFS on Tree

Introduction

Vanilla BFS

Matrix as Graph

Implicit Graph

Directed Graph / Topological Sort

Weighted Graph

Minimum Spanning Tree

Speedrun

Introduction

Same Direction

Opposite Direction

Sliding Window

Prefix Sum

Cycle Finding

Advanced

Speedrun

Introduction

Top K

Moving Best

Multiple Heaps

Introduction

Grid

Dynamic number of subproblems

Interval

Dual-Sequence

Knapsack

Bitmask

Additional Practices

Disjoint Set Union | Union Find

Trie

Data Structure Design

Segment Tree

Interval

Monotonic Stack

Divide and Conquer

Line Sweep

Tree Traversal without Recursion

Greedy

Math

Matrix

Classics

Concepts

Projects

Amazon OA

Microsoft OA

Google OA

Twitter OA

Editorials

2143. Choose Numbers From Two Arrays in Range

Problem Description

Intuition

Solution Approach

Example Walkthrough

Solution Implementation

Time and Space Complexity

Time Complexity

Space Complexity

Recommended Readings