LeetCode #1711 — MEDIUM

Count Good Meals

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

The Problem

Problem Statement

A good meal is a meal that contains exactly two different food items with a sum of deliciousness equal to a power of two.

You can pick any two different foods to make a good meal.

Given an array of integers deliciousness where deliciousness[i] is the deliciousness of the i^th item of food, return the number of different good meals you can make from this list modulo 10⁹ + 7.

Note that items with different indices are considered different even if they have the same deliciousness value.

Example 1:

Input: deliciousness = [1,3,5,7,9]
Output: 4
Explanation: The good meals are (1,3), (1,7), (3,5) and, (7,9).
Their respective sums are 4, 8, 8, and 16, all of which are powers of 2.

Example 2:

Input: deliciousness = [1,1,1,3,3,3,7]
Output: 15
Explanation: The good meals are (1,1) with 3 ways, (1,3) with 9 ways, and (1,7) with 3 ways.

Constraints:

1 <= deliciousness.length <= 10⁵
0 <= deliciousness[i] <= 2²⁰

Step 01

Brute Force Baseline

Problem summary: A good meal is a meal that contains exactly two different food items with a sum of deliciousness equal to a power of two. You can pick any two different foods to make a good meal. Given an array of integers deliciousness where deliciousness[i] is the deliciousness of the ith item of food, return the number of different good meals you can make from this list modulo 109 + 7. Note that items with different indices are considered different even if they have the same deliciousness value.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[1,3,5,7,9]

Example 2

[1,1,1,3,3,3,7]

Core Insight

What unlocks the optimal approach

Note that the number of powers of 2 is at most 21 so this turns the problem to a classic find the number of pairs that sum to a certain value but for 21 values
You need to use something fasters than the NlogN approach since there is already the log of iterating over the powers so one idea is two pointers

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 #1711: Count Good Meals
class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int countPairs(int[] deliciousness) {
        int mx = Arrays.stream(deliciousness).max().getAsInt() << 1;
        int ans = 0;
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int d : deliciousness) {
            for (int s = 1; s <= mx; s <<= 1) {
                ans = (ans + cnt.getOrDefault(s - d, 0)) % MOD;
            }
            cnt.merge(d, 1, Integer::sum);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1711: Count Good Meals
func countPairs(deliciousness []int) (ans int) {
	mx := slices.Max(deliciousness) << 1
	const mod int = 1e9 + 7
	cnt := map[int]int{}
	for _, d := range deliciousness {
		for s := 1; s <= mx; s <<= 1 {
			ans = (ans + cnt[s-d]) % mod
		}
		cnt[d]++
	}
	return
}

# Accepted solution for LeetCode #1711: Count Good Meals
class Solution:
    def countPairs(self, deliciousness: List[int]) -> int:
        mod = 10**9 + 7
        mx = max(deliciousness) << 1
        cnt = Counter()
        ans = 0
        for d in deliciousness:
            s = 1
            while s <= mx:
                ans = (ans + cnt[s - d]) % mod
                s <<= 1
            cnt[d] += 1
        return ans

// Accepted solution for LeetCode #1711: Count Good Meals
struct Solution;

use std::collections::HashMap;

const MOD: i64 = 1_000_000_007;

impl Solution {
    fn count_pairs(deliciousness: Vec<i32>) -> i32 {
        let mut hm: HashMap<i32, i64> = HashMap::new();
        let mut res = 0;
        for x in deliciousness {
            for i in 0..22 {
                let sum = 1 << i;
                let y = sum - x;
                if let Some(k) = hm.get(&y) {
                    res += k;
                    res %= MOD;
                }
            }
            *hm.entry(x).or_default() += 1;
        }
        res as i32
    }
}

#[test]
fn test() {
    let deliciousness = vec![1, 3, 5, 7, 9];
    let res = 4;
    assert_eq!(Solution::count_pairs(deliciousness), res);
    let deliciousness = vec![1, 1, 1, 3, 3, 3, 7];
    let res = 15;
    assert_eq!(Solution::count_pairs(deliciousness), res);
    let deliciousness = vec![
        149, 107, 1, 63, 0, 1, 6867, 1325, 5611, 2581, 39, 89, 46, 18, 12, 20, 22, 234,
    ];
    let res = 12;
    assert_eq!(Solution::count_pairs(deliciousness), res);
}

// Accepted solution for LeetCode #1711: Count Good Meals
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1711: Count Good Meals
// class Solution {
//     private static final int MOD = (int) 1e9 + 7;
// 
//     public int countPairs(int[] deliciousness) {
//         int mx = Arrays.stream(deliciousness).max().getAsInt() << 1;
//         int ans = 0;
//         Map<Integer, Integer> cnt = new HashMap<>();
//         for (int d : deliciousness) {
//             for (int s = 1; s <= mx; s <<= 1) {
//                 ans = (ans + cnt.getOrDefault(s - d, 0)) % MOD;
//             }
//             cnt.merge(d, 1, Integer::sum);
//         }
//         return 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)

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.

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.