Choose Numbers From Two Arrays in Range

Choose Numbers From Two Arrays in Range - Problem

You are given two 0-indexed integer arrays nums1 and nums2 of length n.

A range [l, r] (inclusive) where 0 <= l <= r < n is balanced if:

For every i in the range [l, r], you pick either nums1[i] or nums2[i].
The sum of the numbers you pick from nums1 equals the sum of the numbers you pick from nums2 (the sum is considered to be 0 if you pick no numbers from an array).

Two balanced ranges [l1, r1] and [l2, r2] are considered to be different if at least one of the following is true:

l1 != l2
r1 != r2
nums1[i] is picked in the first range, and nums2[i] is picked in the second range or vice versa for at least one i.

Return the number of different ranges that are balanced. Since the answer may be very large, return it modulo 10⁹ + 7.

Input & Output

Example 1 — Basic Case

$ Input: nums1 = [1,2,3], nums2 = [4,5,6]

› Output: 0

💡 Note: For nums1=[1,2,3], nums2=[4,5,6], checking all ranges: Range [0,0] has 2 selections with sums (1,0) and (0,4) - not balanced. Range [1,1] has 2 selections with sums (2,0) and (0,5) - not balanced. Range [2,2] has 2 selections with sums (3,0) and (0,6) - not balanced. Range [0,1] has 4 selections with sums (3,0), (1,5), (2,4), (0,9) - not balanced. Range [1,2] has 4 selections with sums (5,0), (2,6), (3,5), (0,11) - not balanced. Range [0,2] has 8 selections with sums (6,0), (3,6), (4,5), (1,11), (5,4), (2,10), (3,9), (0,15) - not balanced. Total: 0 balanced ranges.

Example 2 — Equal Arrays

$ Input: nums1 = [1,1], nums2 = [1,1]

› Output: 2

💡 Note: For nums1=[1,1], nums2=[1,1]: Range [0,0] has selections (1,0) and (0,1) - not balanced. Range [1,1] has selections (1,0) and (0,1) - not balanced. Range [0,1] has selections (2,0), (1,1), (1,1), (0,2) where two selections give sum1=1, sum2=1 - these are balanced. Total: 2 balanced ranges.

Example 3 — Single Element

$ Input: nums1 = [5], nums2 = [5]

› Output: 0

💡 Note: For nums1=[5], nums2=[5]: Only range [0,0] with selections (5,0) and (0,5) - neither balanced. Total: 0 balanced ranges.

Constraints

1 ≤ n ≤ 100
-1000 ≤ nums1[i], nums2[i] ≤ 1000
Answer fits in 32-bit integer after modulo

Visualization

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Asked in

G Google 15 f Meta 12 a Amazon 8

The key insight is to transform the problem by creating a difference array diff[i] = nums1[i] - nums2[i], then use dynamic programming to count ways to achieve sum = 0 for each range. Best approach uses space-optimized DP with Time: O(n² × S), Space: O(S) where S is the sum range.

Common Approaches

✓ Brute Force - Try All Ranges and Selections

⏱️ Time: O(n² × 2ⁿ) Space: O(1)

For each possible range [l,r], try all 2^(r-l+1) ways to pick from nums1 or nums2, and count how many result in equal sums from both arrays.

Dynamic Programming with Memoization

⏱️ Time: O(n² × S) Space: O(n × S)

For each range, use dynamic programming to count the number of ways to achieve zero difference between sums. Transform the problem into counting paths in a difference space.

Optimized DP with Difference Array

⏱️ Time: O(n² × S) Space: O(S)

Convert nums1[i] and nums2[i] into difference array diff[i] = nums1[i] - nums2[i]. Now we need to count ways to select subset with sum = 0 for each range.

Brute Force - Try All Ranges and Selections — Algorithm Steps

Step 1: Iterate through all possible ranges [l,r] where 0 <= l <= r < n
Step 2: For each range, generate all 2^(range_size) possible selections using bit manipulation
Step 3: For each selection, calculate sum from nums1 and nums2, check if equal
Step 4: Count all balanced selections and return total modulo 10^9 + 7

Visualization

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Step-by-Step Walkthrough

Generate Ranges

Try all [l,r] ranges

Generate Selections

For each range, try all 2^k ways to pick

Count Balanced

Count selections where sum1 = sum2

Code -

solution.c — C

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

int solution(int* nums1, int* nums2, int n) {
    const int MOD = 1000000007;
    int result = 0;
    
    for (int l = 0; l < n; l++) {
        for (int r = l; r < n; r++) {
            int rangeSize = r - l + 1;
            // Try all 2^rangeSize selections
            for (int mask = 0; mask < (1 << rangeSize); mask++) {
                int sum1 = 0, sum2 = 0;
                for (int i = 0; i < rangeSize; i++) {
                    if (mask & (1 << i)) {
                        sum1 += nums1[l + i];
                    } else {
                        sum2 += nums2[l + i];
                    }
                }
                if (sum1 == sum2) {
                    result = (result + 1) % MOD;
                }
            }
        }
    }
    
    return result;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse first array
    int nums1[100], n1 = 0;
    char* token = strtok(line, "[,]");
    while (token) {
        if (token[0] >= '0' || token[0] == '-') {
            nums1[n1++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    fgets(line, sizeof(line), stdin);
    int nums2[100], n2 = 0;
    token = strtok(line, "[,]");
    while (token) {
        if (token[0] >= '0' || token[0] == '-') {
            nums2[n2++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    printf("%d\n", solution(nums1, nums2, n1));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n² × 2ⁿ)

O(n²) ranges, each with O(2^n) selections to check in worst case

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for counting, no extra data structures

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Ranges and Selections — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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