LeetCode #3117 — HARD

Minimum Sum of Values by Dividing Array

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given two arrays nums and andValues of length n and m respectively.

The value of an array is equal to the last element of that array.

You have to divide nums into m disjoint contiguous subarrays such that for the i^th subarray [l_i, r_i], the bitwise AND of the subarray elements is equal to andValues[i], in other words, nums[l_i] & nums[l_i + 1] & ... & nums[r_i] == andValues[i] for all 1 <= i <= m, where & represents the bitwise AND operator.

Return the minimum possible sum of the values of the m subarrays nums is divided into. If it is not possible to divide nums into m subarrays satisfying these conditions, return -1.

Example 1:

Input: nums = [1,4,3,3,2], andValues = [0,3,3,2]

Output: 12

Explanation:

The only possible way to divide nums is:

[1,4] as 1 & 4 == 0.
[3] as the bitwise AND of a single element subarray is that element itself.
[3] as the bitwise AND of a single element subarray is that element itself.
[2] as the bitwise AND of a single element subarray is that element itself.

The sum of the values for these subarrays is 4 + 3 + 3 + 2 = 12.

Example 2:

Input: nums = [2,3,5,7,7,7,5], andValues = [0,7,5]

Output: 17

Explanation:

There are three ways to divide nums:

[[2,3,5],[7,7,7],[5]] with the sum of the values 5 + 7 + 5 == 17.
[[2,3,5,7],[7,7],[5]] with the sum of the values 7 + 7 + 5 == 19.
[[2,3,5,7,7],[7],[5]] with the sum of the values 7 + 7 + 5 == 19.

The minimum possible sum of the values is 17.

Example 3:

Input: nums = [1,2,3,4], andValues = [2]

Output: -1

Explanation:

The bitwise AND of the entire array nums is 0. As there is no possible way to divide nums into a single subarray to have the bitwise AND of elements 2, return -1.

Constraints:

1 <= n == nums.length <= 10⁴
1 <= m == andValues.length <= min(n, 10)
1 <= nums[i] < 10⁵
0 <= andValues[j] < 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given two arrays nums and andValues of length n and m respectively. The value of an array is equal to the last element of that array. You have to divide nums into m disjoint contiguous subarrays such that for the ith subarray [li, ri], the bitwise AND of the subarray elements is equal to andValues[i], in other words, nums[li] & nums[li + 1] & ... & nums[ri] == andValues[i] for all 1 <= i <= m, where & represents the bitwise AND operator. Return the minimum possible sum of the values of the m subarrays nums is divided into. If it is not possible to divide nums into m subarrays satisfying these conditions, return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Dynamic Programming · Bit Manipulation · Segment Tree

Example 1

[1,4,3,3,2]
[0,3,3,2]

Example 2

[2,3,5,7,7,7,5]
[0,7,5]

Example 3

[1,2,3,4]
[2]

Core Insight

What unlocks the optimal approach

Let <code>dp[i][j]</code> be the optimal answer to split <code>nums[0..(i - 1)]</code> into the first <code>j</code> andValues.
<code>dp[i][j] = min(dp[(i - z)][j - 1]) + nums[i - 1]</code> over all <code>x <= z <= y</code> and <code>dp[0][0] = 0</code>, where <code>x</code> and <code>y</code> are the longest and shortest subarrays ending with <code>nums[i - 1]</code> and the bitwise-and of all the values in it is <code>andValues[j - 1]</code>.
The answer is <code>dp[n][m]</code>.
To calculate <code>x</code> and <code>y</code>, we can use binary search (or sliding window). Note that the more values we have, the smaller the <code>AND</code> value is.
To calculate the result, we need to support RMQ (range minimum query). Segment tree is one way to do it in <code>O(log(n))</code>. But we can use Monotonic Queue since the ranges are indeed “sliding to right” which can be reduced to the classical minimum value in sliding window problem, for a <code>O(n)</code> solution.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3117: Minimum Sum of Values by Dividing Array
class Solution {
    private int[] nums;
    private int[] andValues;
    private final int inf = 1 << 29;
    private Map<Long, Integer> f = new HashMap<>();

    public int minimumValueSum(int[] nums, int[] andValues) {
        this.nums = nums;
        this.andValues = andValues;
        int ans = dfs(0, 0, -1);
        return ans >= inf ? -1 : ans;
    }

    private int dfs(int i, int j, int a) {
        if (nums.length - i < andValues.length - j) {
            return inf;
        }
        if (j == andValues.length) {
            return i == nums.length ? 0 : inf;
        }
        a &= nums[i];
        if (a < andValues[j]) {
            return inf;
        }
        long key = (long) i << 36 | (long) j << 32 | a;
        if (f.containsKey(key)) {
            return f.get(key);
        }

        int ans = dfs(i + 1, j, a);
        if (a == andValues[j]) {
            ans = Math.min(ans, dfs(i + 1, j + 1, -1) + nums[i]);
        }
        f.put(key, ans);
        return ans;
    }
}

// Accepted solution for LeetCode #3117: Minimum Sum of Values by Dividing Array
func minimumValueSum(nums []int, andValues []int) int {
	n, m := len(nums), len(andValues)
	f := map[int]int{}
	const inf int = 1 << 29
	var dfs func(i, j, a int) int
	dfs = func(i, j, a int) int {
		if n-i < m-j {
			return inf
		}
		if j == m {
			if i == n {
				return 0
			}
			return inf
		}
		a &= nums[i]
		if a < andValues[j] {
			return inf
		}
		key := i<<36 | j<<32 | a
		if v, ok := f[key]; ok {
			return v
		}
		ans := dfs(i+1, j, a)
		if a == andValues[j] {
			ans = min(ans, dfs(i+1, j+1, -1)+nums[i])
		}
		f[key] = ans
		return ans
	}
	if ans := dfs(0, 0, -1); ans < inf {
		return ans
	}
	return -1
}

# Accepted solution for LeetCode #3117: Minimum Sum of Values by Dividing Array
class Solution:
    def minimumValueSum(self, nums: List[int], andValues: List[int]) -> int:
        @cache
        def dfs(i: int, j: int, a: int) -> int:
            if n - i < m - j:
                return inf
            if j == m:
                return 0 if i == n else inf
            a &= nums[i]
            if a < andValues[j]:
                return inf
            ans = dfs(i + 1, j, a)
            if a == andValues[j]:
                ans = min(ans, dfs(i + 1, j + 1, -1) + nums[i])
            return ans

        n, m = len(nums), len(andValues)
        ans = dfs(0, 0, -1)
        return ans if ans < inf else -1

// Accepted solution for LeetCode #3117: Minimum Sum of Values by Dividing Array
/**
 * [3117] Minimum Sum of Values by Dividing Array
 */
pub struct Solution {}

// submission codes start here
use std::collections::HashMap;

const INFINITE: i32 = (1 << 20) - 1;

impl Solution {
    pub fn minimum_value_sum(nums: Vec<i32>, and_values: Vec<i32>) -> i32 {
        let (n, m) = (nums.len(), and_values.len());
        let mut memory = vec![HashMap::new(); m * n];

        let result = Self::dfs(&mut memory, 0, 0, INFINITE, &nums, &and_values);
        if result == INFINITE {
            -1
        } else {
            result
        }
    }

    fn dfs(
        memory: &mut Vec<HashMap<i32, i32>>,
        i: usize,
        j: usize,
        current: i32,
        nums: &Vec<i32>,
        and_values: &Vec<i32>,
    ) -> i32 {
        let (n, m) = (nums.len(), and_values.len());
        let key = m * i + j;

        if i == n && j == m {
            return 0;
        }
        if i == n || j == m {
            return INFINITE;
        }
        if let Some(r) = memory[key].get(&current) {
            return *r;
        }

        let current = current & nums[i];
        if current & and_values[j] < and_values[j] {
            return INFINITE;
        }

        let mut result = Self::dfs(memory, i + 1, j, current, nums, and_values);
        if current == and_values[j] {
            result =
                result.min(Self::dfs(memory, i + 1, j + 1, INFINITE, nums, and_values) + nums[i]);
        }

        memory[key].insert(current, result);
        return result;
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3117() {
        assert_eq!(
            12,
            Solution::minimum_value_sum(vec![1, 4, 3, 3, 2], vec![0, 3, 3, 2])
        );
    }
}

// Accepted solution for LeetCode #3117: Minimum Sum of Values by Dividing Array
function minimumValueSum(nums: number[], andValues: number[]): number {
    const [n, m] = [nums.length, andValues.length];
    const f: Map<bigint, number> = new Map();
    const dfs = (i: number, j: number, a: number): number => {
        if (n - i < m - j) {
            return Infinity;
        }
        if (j === m) {
            return i === n ? 0 : Infinity;
        }
        a &= nums[i];
        if (a < andValues[j]) {
            return Infinity;
        }
        const key = (BigInt(i) << 36n) | (BigInt(j) << 32n) | BigInt(a);
        if (f.has(key)) {
            return f.get(key)!;
        }
        let ans = dfs(i + 1, j, a);
        if (a === andValues[j]) {
            ans = Math.min(ans, dfs(i + 1, j + 1, -1) + nums[i]);
        }
        f.set(key, ans);
        return ans;
    };
    const ans = dfs(0, 0, -1);
    return ans >= Infinity ? -1 : ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m × log M)

Space

O(n × m × log M)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Minimum Sum of Values by Dividing Array

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

Boundary update without `+1` / `-1`

State misses one required dimension