LeetCode #3770 — MEDIUM

Largest Prime from Consecutive Prime Sum

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

The Problem

Problem Statement

You are given an integer n.

Return the largest prime number less than or equal to n that can be expressed as the sum of one or more consecutive prime numbers starting from 2. If no such number exists, return 0.

Example 1:

Input: n = 20

Output: 17

Explanation:

The prime numbers less than or equal to n = 20 which are consecutive prime sums are:

2 = 2
5 = 2 + 3
17 = 2 + 3 + 5 + 7

The largest is 17, so it is the answer.

Example 2:

Input: n = 2

Output: 2

Explanation:

The only consecutive prime sum less than or equal to 2 is 2 itself.

Constraints:

1 <= n <= 5 * 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer n. Return the largest prime number less than or equal to n that can be expressed as the sum of one or more consecutive prime numbers starting from 2. If no such number exists, return 0.

Baseline thinking

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

Pattern signal: Array · Math

Example 1

Example 2

Step 02

Core Insight

What unlocks the optimal approach

Generate all prime numbers up to <code>n</code> (use sieve or trial division).
Compute consecutive sums starting from 2 until the sum exceeds <code>n</code>.
Find the largest sum that is also a prime.

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 #3770: Largest Prime from Consecutive Prime Sum
class Solution {
    private static final int MX = 500000;
    private static final boolean[] IS_PRIME = new boolean[MX + 1];
    private static final List<Integer> PRIMES = new ArrayList<>();
    private static final List<Integer> S = new ArrayList<>();

    static {
        Arrays.fill(IS_PRIME, true);
        IS_PRIME[0] = false;
        IS_PRIME[1] = false;

        for (int i = 2; i <= MX; i++) {
            if (IS_PRIME[i]) {
                PRIMES.add(i);
                if ((long) i * i <= MX) {
                    for (int j = i * i; j <= MX; j += i) {
                        IS_PRIME[j] = false;
                    }
                }
            }
        }

        S.add(0);
        int t = 0;
        for (int x : PRIMES) {
            t += x;
            if (t > MX) {
                break;
            }
            if (IS_PRIME[t]) {
                S.add(t);
            }
        }
    }

    public int largestPrime(int n) {
        int i = Collections.binarySearch(S, n + 1);
        if (i < 0) {
            i = ~i;
        }
        return S.get(i - 1);
    }
}

// Accepted solution for LeetCode #3770: Largest Prime from Consecutive Prime Sum
const MX = 500000

var (
	isPrime = make([]bool, MX+1)
	primes  []int
	S       []int
)

func init() {
	for i := range isPrime {
		isPrime[i] = true
	}
	isPrime[0] = false
	isPrime[1] = false

	for i := 2; i <= MX; i++ {
		if isPrime[i] {
			primes = append(primes, i)
			if i*i <= MX {
				for j := i * i; j <= MX; j += i {
					isPrime[j] = false
				}
			}
		}
	}

	S = append(S, 0)
	t := 0
	for _, x := range primes {
		t += x
		if t > MX {
			break
		}
		if isPrime[t] {
			S = append(S, t)
		}
	}
}

func largestPrime(n int) int {
	i := sort.SearchInts(S, n+1)
	return S[i-1]
}

# Accepted solution for LeetCode #3770: Largest Prime from Consecutive Prime Sum
mx = 500000
is_prime = [True] * (mx + 1)
is_prime[0] = is_prime[1] = False
primes = []
for i in range(2, mx + 1):
    if is_prime[i]:
        primes.append(i)
        for j in range(i * i, mx + 1, i):
            is_prime[j] = False
s = [0]
t = 0
for x in primes:
    t += x
    if t > mx:
        break
    if is_prime[t]:
        s.append(t)


class Solution:
    def largestPrime(self, n: int) -> int:
        i = bisect_right(s, n) - 1
        return s[i]

// Accepted solution for LeetCode #3770: Largest Prime from Consecutive Prime Sum
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3770: Largest Prime from Consecutive Prime Sum
// class Solution {
//     private static final int MX = 500000;
//     private static final boolean[] IS_PRIME = new boolean[MX + 1];
//     private static final List<Integer> PRIMES = new ArrayList<>();
//     private static final List<Integer> S = new ArrayList<>();
// 
//     static {
//         Arrays.fill(IS_PRIME, true);
//         IS_PRIME[0] = false;
//         IS_PRIME[1] = false;
// 
//         for (int i = 2; i <= MX; i++) {
//             if (IS_PRIME[i]) {
//                 PRIMES.add(i);
//                 if ((long) i * i <= MX) {
//                     for (int j = i * i; j <= MX; j += i) {
//                         IS_PRIME[j] = false;
//                     }
//                 }
//             }
//         }
// 
//         S.add(0);
//         int t = 0;
//         for (int x : PRIMES) {
//             t += x;
//             if (t > MX) {
//                 break;
//             }
//             if (IS_PRIME[t]) {
//                 S.add(t);
//             }
//         }
//     }
// 
//     public int largestPrime(int n) {
//         int i = Collections.binarySearch(S, n + 1);
//         if (i < 0) {
//             i = ~i;
//         }
//         return S.get(i - 1);
//     }
// }

// Accepted solution for LeetCode #3770: Largest Prime from Consecutive Prime Sum
const MX = 500000;

const isPrime: boolean[] = Array(MX + 1).fill(true);
isPrime[0] = false;
isPrime[1] = false;

const primes: number[] = [];
const s: number[] = [];

(function init() {
    for (let i = 2; i <= MX; i++) {
        if (isPrime[i]) {
            primes.push(i);
            if (i * i <= MX) {
                for (let j = i * i; j <= MX; j += i) {
                    isPrime[j] = false;
                }
            }
        }
    }

    s.push(0);
    let t = 0;
    for (const x of primes) {
        t += x;
        if (t > MX) break;
        if (isPrime[t]) {
            s.push(t);
        }
    }
})();

function largestPrime(n: number): number {
    const i = _.sortedIndex(s, n + 1) - 1;
    return s[i];
}

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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.