LeetCode #3556 — MEDIUM

Sum of Largest Prime Substrings

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

The Problem

Problem Statement

Given a string s, find the sum of the 3 largest unique prime numbers that can be formed using any of its substrings.

Return the sum of the three largest unique prime numbers that can be formed. If fewer than three exist, return the sum of all available primes. If no prime numbers can be formed, return 0.

Note: Each prime number should be counted only once, even if it appears in multiple substrings. Additionally, when converting a substring to an integer, any leading zeros are ignored.

Example 1:

Input: s = "12234"

Output: 1469

Explanation:

The unique prime numbers formed from the substrings of "12234" are 2, 3, 23, 223, and 1223.
The 3 largest primes are 1223, 223, and 23. Their sum is 1469.

Example 2:

Input: s = "111"

Output: 11

Explanation:

The unique prime number formed from the substrings of "111" is 11.
Since there is only one prime number, the sum is 11.

Constraints:

1 <= s.length <= 10
s consists of only digits.

Step 01

Brute Force Baseline

Problem summary: Given a string s, find the sum of the 3 largest unique prime numbers that can be formed using any of its substrings. Return the sum of the three largest unique prime numbers that can be formed. If fewer than three exist, return the sum of all available primes. If no prime numbers can be formed, return 0. Note: Each prime number should be counted only once, even if it appears in multiple substrings. Additionally, when converting a substring to an integer, any leading zeros are ignored.

Baseline thinking

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

Pattern signal: Hash Map · Math

Example 1

"12234"

Example 2

"111"

Step 02

Core Insight

What unlocks the optimal approach

Iterate over all substrings of <code>s</code> to generate candidate numbers.
Check each candidate for primality in <code>O(sqrt(n))</code> time.
Store unique primes, then sum the three largest (or all if fewer than three).

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 #3556: Sum of Largest Prime Substrings
class Solution {
    public long sumOfLargestPrimes(String s) {
        Set<Long> st = new HashSet<>();
        int n = s.length();

        for (int i = 0; i < n; i++) {
            long x = 0;
            for (int j = i; j < n; j++) {
                x = x * 10 + (s.charAt(j) - '0');
                if (is_prime(x)) {
                    st.add(x);
                }
            }
        }

        List<Long> sorted = new ArrayList<>(st);
        Collections.sort(sorted);

        long ans = 0;
        int start = Math.max(0, sorted.size() - 3);
        for (int idx = start; idx < sorted.size(); idx++) {
            ans += sorted.get(idx);
        }
        return ans;
    }

    private boolean is_prime(long x) {
        if (x < 2) return false;
        for (long i = 2; i * i <= x; i++) {
            if (x % i == 0) return false;
        }
        return true;
    }
}

// Accepted solution for LeetCode #3556: Sum of Largest Prime Substrings
func sumOfLargestPrimes(s string) (ans int64) {
	st := make(map[int64]struct{})
	n := len(s)

	for i := 0; i < n; i++ {
		var x int64 = 0
		for j := i; j < n; j++ {
			x = x*10 + int64(s[j]-'0')
			if isPrime(x) {
				st[x] = struct{}{}
			}
		}
	}

	nums := make([]int64, 0, len(st))
	for num := range st {
		nums = append(nums, num)
	}
	sort.Slice(nums, func(i, j int) bool { return nums[i] < nums[j] })
	for i := len(nums) - 1; i >= 0 && len(nums)-i <= 3; i-- {
		ans += nums[i]
	}
	return
}

func isPrime(x int64) bool {
	if x < 2 {
		return false
	}
	sqrtX := int64(math.Sqrt(float64(x)))
	for i := int64(2); i <= sqrtX; i++ {
		if x%i == 0 {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #3556: Sum of Largest Prime Substrings
class Solution:
    def sumOfLargestPrimes(self, s: str) -> int:
        def is_prime(x: int) -> bool:
            if x < 2:
                return False
            return all(x % i for i in range(2, int(sqrt(x)) + 1))

        st = set()
        n = len(s)
        for i in range(n):
            x = 0
            for j in range(i, n):
                x = x * 10 + int(s[j])
                if is_prime(x):
                    st.add(x)
        return sum(sorted(st)[-3:])

// Accepted solution for LeetCode #3556: Sum of Largest Prime Substrings
// 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 #3556: Sum of Largest Prime Substrings
// class Solution {
//     public long sumOfLargestPrimes(String s) {
//         Set<Long> st = new HashSet<>();
//         int n = s.length();
// 
//         for (int i = 0; i < n; i++) {
//             long x = 0;
//             for (int j = i; j < n; j++) {
//                 x = x * 10 + (s.charAt(j) - '0');
//                 if (is_prime(x)) {
//                     st.add(x);
//                 }
//             }
//         }
// 
//         List<Long> sorted = new ArrayList<>(st);
//         Collections.sort(sorted);
// 
//         long ans = 0;
//         int start = Math.max(0, sorted.size() - 3);
//         for (int idx = start; idx < sorted.size(); idx++) {
//             ans += sorted.get(idx);
//         }
//         return ans;
//     }
// 
//     private boolean is_prime(long x) {
//         if (x < 2) return false;
//         for (long i = 2; i * i <= x; i++) {
//             if (x % i == 0) return false;
//         }
//         return true;
//     }
// }

// Accepted solution for LeetCode #3556: Sum of Largest Prime Substrings
function sumOfLargestPrimes(s: string): number {
    const st = new Set<number>();
    const n = s.length;

    for (let i = 0; i < n; i++) {
        let x = 0;
        for (let j = i; j < n; j++) {
            x = x * 10 + Number(s[j]);
            if (isPrime(x)) {
                st.add(x);
            }
        }
    }

    const sorted = Array.from(st).sort((a, b) => a - b);
    const topThree = sorted.slice(-3);
    return topThree.reduce((sum, val) => sum + val, 0);
}

function isPrime(x: number): boolean {
    if (x < 2) return false;
    for (let i = 2; i * i <= x; i++) {
        if (x % i === 0) return false;
    }
    return true;
}

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^2 × sqrtM)

Space

O(n^2)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

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.