LeetCode #991 — MEDIUM

Broken Calculator

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

The Problem

Problem Statement

There is a broken calculator that has the integer startValue on its display initially. In one operation, you can:

multiply the number on display by 2, or
subtract 1 from the number on display.

Given two integers startValue and target, return the minimum number of operations needed to display target on the calculator.

Example 1:

Input: startValue = 2, target = 3
Output: 2
Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.

Example 2:

Input: startValue = 5, target = 8
Output: 2
Explanation: Use decrement and then double {5 -> 4 -> 8}.

Example 3:

Input: startValue = 3, target = 10
Output: 3
Explanation: Use double, decrement and double {3 -> 6 -> 5 -> 10}.

Constraints:

1 <= startValue, target <= 10⁹

Step 01

Brute Force Baseline

Problem summary: There is a broken calculator that has the integer startValue on its display initially. In one operation, you can: multiply the number on display by 2, or subtract 1 from the number on display. Given two integers startValue and target, return the minimum number of operations needed to display target on the calculator.

Baseline thinking

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

Pattern signal: Math · Greedy

Example 1

2
3

Example 2

5
8

Example 3

3
10

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #991: Broken Calculator
class Solution {
    public int brokenCalc(int startValue, int target) {
        int ans = 0;
        while (startValue < target) {
            if ((target & 1) == 1) {
                target++;
            } else {
                target >>= 1;
            }
            ans += 1;
        }
        ans += startValue - target;
        return ans;
    }
}

// Accepted solution for LeetCode #991: Broken Calculator
func brokenCalc(startValue int, target int) (ans int) {
	for startValue < target {
		if target&1 == 1 {
			target++
		} else {
			target >>= 1
		}
		ans++
	}
	ans += startValue - target
	return
}

# Accepted solution for LeetCode #991: Broken Calculator
class Solution:
    def brokenCalc(self, startValue: int, target: int) -> int:
        ans = 0
        while startValue < target:
            if target & 1:
                target += 1
            else:
                target >>= 1
            ans += 1
        ans += startValue - target
        return ans

// Accepted solution for LeetCode #991: Broken Calculator
struct Solution;

impl Solution {
    fn broken_calc(x: i32, mut y: i32) -> i32 {
        let mut res = 0;
        while y > x {
            if y % 2 == 0 {
                y /= 2;
            } else {
                y += 1;
            }
            res += 1;
        }
        res + x - y
    }
}

#[test]
fn test() {
    let x = 2;
    let y = 3;
    let res = 2;
    assert_eq!(Solution::broken_calc(x, y), res);
    let x = 5;
    let y = 8;
    let res = 2;
    assert_eq!(Solution::broken_calc(x, y), res);
    let x = 3;
    let y = 10;
    let res = 3;
    assert_eq!(Solution::broken_calc(x, y), res);
    let x = 1024;
    let y = 1;
    let res = 1023;
    assert_eq!(Solution::broken_calc(x, y), res);
}

// Accepted solution for LeetCode #991: Broken Calculator
function brokenCalc(startValue: number, target: number): number {
    let ans = 0;
    for (; startValue < target; ++ans) {
        if (target & 1) {
            ++target;
        } else {
            target >>= 1;
        }
    }
    ans += startValue - target;
    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 log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.