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.
Move from brute-force thinking to an efficient approach using math strategy.
Given a callable function f(x, y) with a hidden formula and a value z, reverse engineer the formula and return all positive integer pairs x and y where f(x,y) == z. You may return the pairs in any order.
While the exact formula is hidden, the function is monotonically increasing, i.e.:
f(x, y) < f(x + 1, y)f(x, y) < f(x, y + 1)The function interface is defined like this:
interface CustomFunction {
public:
// Returns some positive integer f(x, y) for two positive integers x and y based on a formula.
int f(int x, int y);
};
We will judge your solution as follows:
9 hidden implementations of CustomFunction, along with a way to generate an answer key of all valid pairs for a specific z.function_id (to determine which implementation to test your code with), and the target z.findSolution and compare your results with the answer key.Accepted.Example 1:
Input: function_id = 1, z = 5 Output: [[1,4],[2,3],[3,2],[4,1]] Explanation: The hidden formula for function_id = 1 is f(x, y) = x + y. The following positive integer values of x and y make f(x, y) equal to 5: x=1, y=4 -> f(1, 4) = 1 + 4 = 5. x=2, y=3 -> f(2, 3) = 2 + 3 = 5. x=3, y=2 -> f(3, 2) = 3 + 2 = 5. x=4, y=1 -> f(4, 1) = 4 + 1 = 5.
Example 2:
Input: function_id = 2, z = 5 Output: [[1,5],[5,1]] Explanation: The hidden formula for function_id = 2 is f(x, y) = x * y. The following positive integer values of x and y make f(x, y) equal to 5: x=1, y=5 -> f(1, 5) = 1 * 5 = 5. x=5, y=1 -> f(5, 1) = 5 * 1 = 5.
Constraints:
1 <= function_id <= 91 <= z <= 100f(x, y) == z will be in the range 1 <= x, y <= 1000.f(x, y) will fit in 32 bit signed integer if 1 <= x, y <= 1000.Problem summary: Given a callable function f(x, y) with a hidden formula and a value z, reverse engineer the formula and return all positive integer pairs x and y where f(x,y) == z. You may return the pairs in any order. While the exact formula is hidden, the function is monotonically increasing, i.e.: f(x, y) < f(x + 1, y) f(x, y) < f(x, y + 1) The function interface is defined like this: interface CustomFunction { public: // Returns some positive integer f(x, y) for two positive integers x and y based on a formula. int f(int x, int y); }; We will judge your solution as follows: The judge has a list of 9 hidden implementations of CustomFunction, along with a way to generate an answer key of all valid pairs for a specific z. The judge will receive two inputs: a function_id (to determine which implementation to test your code with), and the target z. The judge will call your findSolution and compare your
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Two Pointers · Binary Search
1 5
2 5
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1237: Find Positive Integer Solution for a Given Equation
/*
* // This is the custom function interface.
* // You should not implement it, or speculate about its implementation
* class CustomFunction {
* // Returns f(x, y) for any given positive integers x and y.
* // Note that f(x, y) is increasing with respect to both x and y.
* // i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
* public int f(int x, int y);
* };
*/
class Solution {
public List<List<Integer>> findSolution(CustomFunction customfunction, int z) {
List<List<Integer>> ans = new ArrayList<>();
for (int x = 1; x <= 1000; ++x) {
int l = 1, r = 1000;
while (l < r) {
int mid = (l + r) >> 1;
if (customfunction.f(x, mid) >= z) {
r = mid;
} else {
l = mid + 1;
}
}
if (customfunction.f(x, l) == z) {
ans.add(Arrays.asList(x, l));
}
}
return ans;
}
}
// Accepted solution for LeetCode #1237: Find Positive Integer Solution for a Given Equation
/**
* This is the declaration of customFunction API.
* @param x int
* @param x int
* @return Returns f(x, y) for any given positive integers x and y.
* Note that f(x, y) is increasing with respect to both x and y.
* i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
*/
func findSolution(customFunction func(int, int) int, z int) (ans [][]int) {
for x := 1; x <= 1000; x++ {
y := 1 + sort.Search(999, func(y int) bool { return customFunction(x, y+1) >= z })
if customFunction(x, y) == z {
ans = append(ans, []int{x, y})
}
}
return
}
# Accepted solution for LeetCode #1237: Find Positive Integer Solution for a Given Equation
"""
This is the custom function interface.
You should not implement it, or speculate about its implementation
class CustomFunction:
# Returns f(x, y) for any given positive integers x and y.
# Note that f(x, y) is increasing with respect to both x and y.
# i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
def f(self, x, y):
"""
class Solution:
def findSolution(self, customfunction: "CustomFunction", z: int) -> List[List[int]]:
ans = []
for x in range(1, z + 1):
y = 1 + bisect_left(
range(1, z + 1), z, key=lambda y: customfunction.f(x, y)
)
if customfunction.f(x, y) == z:
ans.append([x, y])
return ans
// Accepted solution for LeetCode #1237: Find Positive Integer Solution for a Given Equation
struct Solution;
struct CustomFunction {
fp: fn(i32, i32) -> i32,
}
impl CustomFunction {
fn f(&self, x: i32, y: i32) -> i32 {
(self.fp)(x, y)
}
}
impl Solution {
fn find_solution(customfunction: &CustomFunction, z: i32) -> Vec<Vec<i32>> {
let mut res = vec![];
for i in 0..1000 {
for j in 0..1000 {
let x = (i + 1) as i32;
let y = (j + 1) as i32;
if customfunction.f(x, y) == z {
res.push(vec![x, y]);
}
}
}
res
}
}
#[test]
fn test() {
let cf = CustomFunction { fp: |x, y| x + y };
let z = 5;
let res = vec_vec_i32![[1, 4], [2, 3], [3, 2], [4, 1]];
assert_eq!(Solution::find_solution(&cf, z), res);
let cf = CustomFunction { fp: |x, y| x * y };
let z = 5;
let res = vec_vec_i32![[1, 5], [5, 1]];
assert_eq!(Solution::find_solution(&cf, z), res);
}
// Accepted solution for LeetCode #1237: Find Positive Integer Solution for a Given Equation
/**
* // This is the CustomFunction's API interface.
* // You should not implement it, or speculate about its implementation
* class CustomFunction {
* f(x: number, y: number): number {}
* }
*/
function findSolution(customfunction: CustomFunction, z: number): number[][] {
const ans: number[][] = [];
for (let x = 1; x <= 1000; ++x) {
let l = 1;
let r = 1000;
while (l < r) {
const mid = (l + r) >> 1;
if (customfunction.f(x, mid) >= z) {
r = mid;
} else {
l = mid + 1;
}
}
if (customfunction.f(x, l) == z) {
ans.push([x, l]);
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.
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.