LeetCode #1477 — MEDIUM

Find Two Non-overlapping Sub-arrays Each With Target Sum

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given an array of integers arr and an integer target.

You have to find two non-overlapping sub-arrays of arr each with a sum equal target. There can be multiple answers so you have to find an answer where the sum of the lengths of the two sub-arrays is minimum.

Return the minimum sum of the lengths of the two required sub-arrays, or return -1 if you cannot find such two sub-arrays.

Example 1:

Input: arr = [3,2,2,4,3], target = 3
Output: 2
Explanation: Only two sub-arrays have sum = 3 ([3] and [3]). The sum of their lengths is 2.

Example 2:

Input: arr = [7,3,4,7], target = 7
Output: 2
Explanation: Although we have three non-overlapping sub-arrays of sum = 7 ([7], [3,4] and [7]), but we will choose the first and third sub-arrays as the sum of their lengths is 2.

Example 3:

Input: arr = [4,3,2,6,2,3,4], target = 6
Output: -1
Explanation: We have only one sub-array of sum = 6.

Constraints:

1 <= arr.length <= 10⁵
1 <= arr[i] <= 1000
1 <= target <= 10⁸

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: You are given an array of integers arr and an integer target. You have to find two non-overlapping sub-arrays of arr each with a sum equal target. There can be multiple answers so you have to find an answer where the sum of the lengths of the two sub-arrays is minimum. Return the minimum sum of the lengths of the two required sub-arrays, or return -1 if you cannot find such two sub-arrays.

Baseline thinking

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

Pattern signal: Array · Hash Map · Binary Search · Dynamic Programming · Sliding Window

Example 1

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

Example 2

[7,3,4,7]
7

Example 3

[4,3,2,6,2,3,4]
6

Related Problems

Find Subarrays With Equal Sum (find-subarrays-with-equal-sum)

Step 02

Core Insight

What unlocks the optimal approach

Let's create two arrays prefix and suffix where prefix[i] is the minimum length of sub-array ends before i and has sum = k, suffix[i] is the minimum length of sub-array starting at or after i and has sum = k.
The answer we are searching for is min(prefix[i] + suffix[i]) for all values of i from 0 to n-1 where n == arr.length.
If you are still stuck with how to build prefix and suffix, you can store for each index i the length of the sub-array starts at i and has sum = k or infinity otherwise, and you can use it to build both prefix and suffix.

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

Upper-end input sizes

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 #1477: Find Two Non-overlapping Sub-arrays Each With Target Sum
class Solution {
    public int minSumOfLengths(int[] arr, int target) {
        Map<Integer, Integer> d = new HashMap<>();
        d.put(0, 0);
        int n = arr.length;
        int[] f = new int[n + 1];
        final int inf = 1 << 30;
        f[0] = inf;
        int s = 0, ans = inf;
        for (int i = 1; i <= n; ++i) {
            int v = arr[i - 1];
            s += v;
            f[i] = f[i - 1];
            if (d.containsKey(s - target)) {
                int j = d.get(s - target);
                f[i] = Math.min(f[i], i - j);
                ans = Math.min(ans, f[j] + i - j);
            }
            d.put(s, i);
        }
        return ans > n ? -1 : ans;
    }
}

// Accepted solution for LeetCode #1477: Find Two Non-overlapping Sub-arrays Each With Target Sum
func minSumOfLengths(arr []int, target int) int {
	d := map[int]int{0: 0}
	const inf = 1 << 30
	s, n := 0, len(arr)
	f := make([]int, n+1)
	f[0] = inf
	ans := inf
	for i, v := range arr {
		i++
		f[i] = f[i-1]
		s += v
		if j, ok := d[s-target]; ok {
			f[i] = min(f[i], i-j)
			ans = min(ans, f[j]+i-j)
		}
		d[s] = i
	}
	if ans > n {
		return -1
	}
	return ans
}

# Accepted solution for LeetCode #1477: Find Two Non-overlapping Sub-arrays Each With Target Sum
class Solution:
    def minSumOfLengths(self, arr: List[int], target: int) -> int:
        d = {0: 0}
        s, n = 0, len(arr)
        f = [inf] * (n + 1)
        ans = inf
        for i, v in enumerate(arr, 1):
            s += v
            f[i] = f[i - 1]
            if s - target in d:
                j = d[s - target]
                f[i] = min(f[i], i - j)
                ans = min(ans, f[j] + i - j)
            d[s] = i
        return -1 if ans > n else ans

// Accepted solution for LeetCode #1477: Find Two Non-overlapping Sub-arrays Each With Target Sum
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn min_sum_of_lengths(arr: Vec<i32>, target: i32) -> i32 {
        let n = arr.len();
        let mut hm: HashMap<i32, i32> = HashMap::new();
        let mut dp: Vec<i32> = vec![std::i32::MAX; n];
        *hm.entry(0).or_default() = -1;
        let mut prev = 0;
        let mut res = std::i32::MAX;
        let mut min = std::i32::MAX;
        for i in 0..n {
            prev += arr[i];
            if let Some(&j) = hm.get(&(prev - target)) {
                if j > -1 && dp[j as usize] != std::i32::MAX {
                    res = res.min(i as i32 - j + dp[j as usize]);
                }
                min = min.min(i as i32 - j);
            }
            dp[i] = min;
            *hm.entry(prev).or_default() = i as i32;
        }

        if res == std::i32::MAX {
            -1
        } else {
            res
        }
    }
}

#[test]
fn test() {
    let arr = vec![3, 2, 2, 4, 3];
    let target = 3;
    let res = 2;
    assert_eq!(Solution::min_sum_of_lengths(arr, target), res);
    let arr = vec![7, 3, 4, 7];
    let target = 7;
    let res = 2;
    assert_eq!(Solution::min_sum_of_lengths(arr, target), res);
    let arr = vec![4, 3, 2, 6, 2, 3, 4];
    let target = 6;
    let res = -1;
    assert_eq!(Solution::min_sum_of_lengths(arr, target), res);
    let arr = vec![5, 5, 4, 4, 5];
    let target = 3;
    let res = -1;
    assert_eq!(Solution::min_sum_of_lengths(arr, target), res);
    let arr = vec![3, 1, 1, 1, 5, 1, 2, 1];
    let target = 3;
    let res = 3;
    assert_eq!(Solution::min_sum_of_lengths(arr, target), res);
    let arr = vec![1, 6, 1];
    let target = 7;
    let res = -1;
    assert_eq!(Solution::min_sum_of_lengths(arr, target), res);
}

// Accepted solution for LeetCode #1477: Find Two Non-overlapping Sub-arrays Each With Target Sum
function minSumOfLengths(arr: number[], target: number): number {
    const d = new Map<number, number>();
    d.set(0, 0);
    let s = 0;
    const n = arr.length;
    const f: number[] = Array(n + 1);
    const inf = 1 << 30;
    f[0] = inf;
    let ans = inf;
    for (let i = 1; i <= n; ++i) {
        const v = arr[i - 1];
        s += v;
        f[i] = f[i - 1];
        if (d.has(s - target)) {
            const j = d.get(s - target)!;
            f[i] = Math.min(f[i], i - j);
            ans = Math.min(ans, f[j] + i - j);
        }
        d.set(s, i);
    }
    return ans > n ? -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(log n)

Space

O(1)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.