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Hi👋SpeedAlgo is developed and maintained by Hi👋WebEnterprise (https://hiwebenterprise.com), an engineering organization focused on developer education, deliberate practice methodologies, and cognitive ergonomics for technical interviews.

How does HiSpeedAlgo prepare candidates for senior and staff SWE interviews?

Rather than rote LeetCode grinding, HiSpeedAlgo enforces deliberate practice: proving invariants, articulating Big-O tradeoffs, managing edge cases, and defending algorithmic approaches to a Senior SWE AI Mock Coach in an integrated Monaco code editor.

6. Minimum Moves to Sort Binary Array • Invariant Proof & Speed-Run

Theory Context•Two Pointers & Sliding Window

HardLC 2193

Minimum Moves to Sort Binary Array (Amazon OA)

Target Frequency:AmazonMicrosoft

Given an array of binary digits, 0 and 1, sort the array so that all zeros are at one end and all ones are at the other. Which end does not matter. To sort the array, swap any two adjacent elements. Determine the minimum number of swaps to sort the array.

Worked Examples

Example 1

Input:arr = [0, 1, 0, 1]

Output:1

Explanation: With 1 move, switching elements 1 and 2, yields [0, 0, 1, 1], a sorted array.

Example 2

Input:arr = [1, 1, 1, 1, 0, 0, 0, 0]

Output:0

Explanation: The array is already sorted with all 1s on the left and all 0s on the right, so 0 moves are necessary.

Example 3

Input:arr = [1, 1, 1, 1, 0, 1, 0, 1]

Output:3

Explanation: Perform the minimal sequence of 3 moves to sort the array into [1, 1, 1, 1, 1, 1, 0, 0].

⚖️Formal Constraints & Bounds

1 <= n <= 10⁵
arr[i] is in the set {0, 1}

Deep-Dive & Conceptual Insights

Why It Works & Core Invariant

Adjacent swaps change inversion count by unit distance: bubbling an element from reader to writer takes exactly reader - writer moves.

Real-World Scenario & Production Applications

Sorting network packet buffers, minimal-energy bit-shuffling in memory controllers, and quantum gate scheduling where adjacent exchanges have unit cost.

Subproblems & Recurrence Decomposition3 Phases

🧩Subproblem 1: Zeros-to-Left Compaction Scan

Skip initial zeros already in place at index 0. For each subsequent 0 discovered at reader, accumulate reader - writer adjacent swaps and advance writer by 1.

Mathematical Recurrence / Code Invariant

zero_left = 0
w = 0
while w < len(arr) and arr[w] == 0: w += 1
for r in range(w, len(arr)):
    if arr[r] == 0:
        zero_left += r - w
        w += 1

Step-by-Step Execution Trace Table

Pass	Reader	`arr[reader]`	Condition / Action	Writer	Delta (`reader - writer`)	Cumulative Moves
0s Left	4	0	Match! Bubble to `writer=0`	1	$4 - 0 = 4$	4
0s Left	6	0	Match! Bubble to `writer=1`	2	$6 - 1 = 5$	9 (Total 0s)
1s Left	5	1	Match! Bubble to `writer=4`	5	$5 - 4 = 1$	1
1s Left	7	1	Match! Bubble to `writer=5`	6	$7 - 5 = 2$	3 (Total 1s)
Final	—	—	Return $\min(9, 3)$	—	—	3

Scroll horizontally to see all columns, or expand to full screen

Core Invariant Specification & Code Shape

Archetype Code Shape

Python / Pseudocode

1	zero_left = ones_left = 0
2	w0 = 0; while w0 < n and arr[w0] == 0: w0 += 1
3	for r in range(w0, n): if arr[r] == 0: zero_left += r - w0; w0 += 1
4	w1 = 0; while w1 < n and arr[w1] == 1: w1 += 1
5	for r in range(w1, n): if arr[r] == 1: ones_left += r - w1; w1 += 1
6	return min(zero_left, ones_left)

Target: Minimum Moves to Sort Binary Array (Amazon OA). Elements already in their final position require 0 moves; advance destination frontier.

Boundary Model: Converging Boundaries [left, right] / Sliding Window [L, R]

Indices step monotonically inward or rightward. Each step permanently eliminates an entire row, column, or infeasible candidate window.

Loop Invariant Termination

while (left < right) for converging pointers; while (right < n) with inner window shrink.

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

Sorting a binary array using only adjacent swaps is equivalent to counting the number of inversions required to separate the 0s and 1s. While general inversion counting on arbitrary sequences requires $O(N \log N)$ with a Fenwick tree or Merge Sort, binary elements afford an elegant $O(N)$ two-pointer reader-writer technique.

🏟️ The Analogy: The Conveyor Belt Sorter

Picture an inspection belt where items of two categories pass by. You have a target collection dock on the far left. The writer pointer marks the next open slot at the dock. The reader pointer walks down the belt. When it discovers a target item, passing it hand-to-hand back to the dock takes exactly reader - writer adjacent swaps. As the item docks, writer advances by 1.

🪄 The Mathematical Harmony / Magic Trick

Code / Blueprint

for reader in range(writer, len(arr)):
    if arr[reader] == target:
        moves += reader - writer
        writer += 1

Because every adjacent swap exchanges two neighboring items, shifting an item leftwards by 1 position decrements its distance to writer by 1 and shifts every intervening item rightwards by 1 position. The relative order of the intervening items is preserved without increasing their future distances!

💡 Summary

By running two linear reader-writer compaction passes—one assuming all 0s move to the left, and one assuming all 1s move to the left—we evaluate both valid sorted states in $2N$ operations, returning $\min(\text{moves}_0, \text{moves}_1)$ in $O(N)$ time and $O(1)$ space.

Assuming 0s must always go to the left: The problem specifies 'Which end does not matter.' Sorting to [1, 1, ..., 0, 0] may require fewer adjacent swaps than [0, 0, ..., 1, 1]. Always compute both directions and take the minimum.
Resetting writer to 0 instead of maintaining frontier: The writer index marks where the next matching element must land. Each time a target element is shifted, incrementing writer += 1 is mandatory to avoid computing swaps to already-occupied slots.
Overcomplicating with Fenwick tree or Merge Sort: Inversion count algorithms for arbitrary values take $O(N \log N)$ time and $O(N)$ space. Because all values are binary {0, 1}, two-pointer reader-writer simulation solves this in $O(N)$ time and $O(1)$ space.

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

You will see how a senior engineer computes minimum adjacent swaps using reader-writer pointer distances instead of heavy inversion algorithms.

Pattern Recognition Signals

The 10-second spot

Binary digits {0, 1} + adjacent swaps only + move all 0s to one end and 1s to other end.

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

Invariant: arr[0..writer-1] holds compacted target prefix. Next target at reader requires exactly reader - writer adjacent swaps to reach writer.

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

Assuming 0s must always move left: the prompt states 'which end does not matter', requiring evaluation of both zeros-left and ones-left configurations.
Over-engineering with Fenwick trees or Merge Sort: arbitrary inversion counting is O(N log N), but binary arrays permit an exact O(N) two-pointer simulation.
Forgetting to skip leading elements already settled at the destination frontier before beginning reader scan.

The 60-Second Interview Pitch

Say this out loud before you type a single line

"Because only adjacent swaps are allowed, moving a target digit from index reader to index writer across intervening digits takes exactly reader - writer moves. I run two linear passes: one computing swaps to bring all 0s to the left, and one computing swaps to bring all 1s to the left. Returning min(zeros_left, ones_left) runs in O(N) time and O(1) space."

So: test both polarities (zeros left vs ones left) using reader - writer compaction distance in 2N steps and return the minimum.

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(N)

Look at the execution steps: The array of length N is traversed twice (once for zeros-left, once for ones-left). Each pass visits each index at most once with O(1) arithmetic. Total time is 2N = O(N).

SPACE COMPLEXITY

O(1) Auxiliary

Look at memory allocations: Only integer pointer variables (writer, reader, zero_left_counter, ones_left_counter) are maintained. No auxiliary arrays or recursive stacks are allocated.

Formal Recurrence Relation

$T(N) = O(N)$

Look at the execution steps: The array of length N is traversed twice (once for zeros-left, once for ones-left). Each pass visits each index at most once with O(1) arithmetic. Total time is 2N = O(N).

Derivation Progression

Zeros-to-Left Compaction Pass

$N \text{ iterations}$

Scans array accumulating reader - writer for all 0s.

Ones-to-Left Compaction Pass

$N \text{ iterations}$

Scans array accumulating reader - writer for all 1s.

Total Operations

$N + N = 2N = O(N)$

Strictly linear time with zero dynamic allocation.

Variable Definitions

$N$

Length of the binary input array

Memory Architecture & Bounds

🟣 Call Stack

O(1) Iterative

🔵 Auxiliary Heap

O(1) No heap allocation

🟢 Output Space

O(1) Scalar integer return

Boundary Best / Worst Cases

Best Case

$O(N)$

Average Case

$O(N)$

Worst Case

$O(N)$

Pointer Invariant Transition Progression

Synthesizing vector architecture diagram...

Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

Triggers: **"binary digits 0 and 1"**, **"swap any two adjacent elements"**, **"all zeros at one end and all ones at the other"**, **"minimum number of swaps"**. The restriction to adjacent swaps means each move changes the inversion count by exactly 1. Moving a target element from index reader to index writer across adjacent elements requires exactly reader - writer swaps.

CONSTRAINTS & BOUNDS

$1 \le n \le 10^5$ , $arr[i] \in \{0, 1\}$ . Requires strictly $O(N)$ time and $O(1)$ auxiliary space.

FAANG PRODUCTION TRAPS & EDGE CASES

Skipping dual-polarity check: the prompt states 'which end does not matter', so evaluating both zeros-to-left and ones-to-left is required to obtain the true minimum.

Core Algorithmic State Invariants

1. Inversion Delta via Reader-Writer Distance

Because each adjacent swap exchanges neighboring indices, moving an element from reader to writer takes exactly reader - writer moves. The reader-writer distance computes the exact inversion count in O(1) time per element without requiring physical array manipulation.

2. Frontier Advancing Compaction Invariant

The writer pointer w tracks the next available destination slot for the target digit. Skipping initial target digits settled at the start requires 0 moves. As each subsequent target digit is discovered, advancing writer by 1 preserves relative ordering of compacted elements.

3. Symmetric Dual-Configuration Minimization

Because grouping all zeros on either the left or right side is permissible, evaluating both zeros-to-left and ones-to-left configurations in two sequential linear passes guarantees optimality in O(N) time and O(1) auxiliary space.

Theory Context•Two Pointers & Sliding Window

HardLC 2193

Minimum Moves to Sort Binary Array (Amazon OA)

Target Frequency:AmazonMicrosoft

Worked Examples

Example 1

Input:arr = [0, 1, 0, 1]

Output:1

Explanation: With 1 move, switching elements 1 and 2, yields [0, 0, 1, 1], a sorted array.

Example 2

Input:arr = [1, 1, 1, 1, 0, 0, 0, 0]

Output:0

Explanation: The array is already sorted with all 1s on the left and all 0s on the right, so 0 moves are necessary.

Example 3

Input:arr = [1, 1, 1, 1, 0, 1, 0, 1]

Output:3

Explanation: Perform the minimal sequence of 3 moves to sort the array into [1, 1, 1, 1, 1, 1, 0, 0].

⚖️Formal Constraints & Bounds

1 <= n <= 10⁵
arr[i] is in the set {0, 1}

Deep-Dive & Conceptual Insights

Why It Works & Core Invariant

Adjacent swaps change inversion count by unit distance: bubbling an element from reader to writer takes exactly reader - writer moves.

Real-World Scenario & Production Applications

Sorting network packet buffers, minimal-energy bit-shuffling in memory controllers, and quantum gate scheduling where adjacent exchanges have unit cost.

Subproblems & Recurrence Decomposition3 Phases

🧩Subproblem 1: Zeros-to-Left Compaction Scan

Skip initial zeros already in place at index 0. For each subsequent 0 discovered at reader, accumulate reader - writer adjacent swaps and advance writer by 1.

Mathematical Recurrence / Code Invariant

zero_left = 0
w = 0
while w < len(arr) and arr[w] == 0: w += 1
for r in range(w, len(arr)):
    if arr[r] == 0:
        zero_left += r - w
        w += 1

Step-by-Step Execution Trace Table

Pass	Reader	`arr[reader]`	Condition / Action	Writer	Delta (`reader - writer`)	Cumulative Moves
0s Left	4	0	Match! Bubble to `writer=0`	1	$4 - 0 = 4$	4
0s Left	6	0	Match! Bubble to `writer=1`	2	$6 - 1 = 5$	9 (Total 0s)
1s Left	5	1	Match! Bubble to `writer=4`	5	$5 - 4 = 1$	1
1s Left	7	1	Match! Bubble to `writer=5`	6	$7 - 5 = 2$	3 (Total 1s)
Final	—	—	Return $\min(9, 3)$	—	—	3

Scroll horizontally to see all columns, or expand to full screen

Core Invariant Specification & Code Shape

Archetype Code Shape

Python / Pseudocode

1	zero_left = ones_left = 0
2	w0 = 0; while w0 < n and arr[w0] == 0: w0 += 1
3	for r in range(w0, n): if arr[r] == 0: zero_left += r - w0; w0 += 1
4	w1 = 0; while w1 < n and arr[w1] == 1: w1 += 1
5	for r in range(w1, n): if arr[r] == 1: ones_left += r - w1; w1 += 1
6	return min(zero_left, ones_left)

Target: Minimum Moves to Sort Binary Array (Amazon OA). Elements already in their final position require 0 moves; advance destination frontier.

Boundary Model: Converging Boundaries [left, right] / Sliding Window [L, R]

Indices step monotonically inward or rightward. Each step permanently eliminates an entire row, column, or infeasible candidate window.

Loop Invariant Termination

while (left < right) for converging pointers; while (right < n) with inner window shrink.

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

🏟️ The Analogy: The Conveyor Belt Sorter

🪄 The Mathematical Harmony / Magic Trick

Code / Blueprint

for reader in range(writer, len(arr)):
    if arr[reader] == target:
        moves += reader - writer
        writer += 1

💡 Summary

Assuming 0s must always go to the left: The problem specifies 'Which end does not matter.' Sorting to [1, 1, ..., 0, 0] may require fewer adjacent swaps than [0, 0, ..., 1, 1]. Always compute both directions and take the minimum.
Resetting writer to 0 instead of maintaining frontier: The writer index marks where the next matching element must land. Each time a target element is shifted, incrementing writer += 1 is mandatory to avoid computing swaps to already-occupied slots.
Overcomplicating with Fenwick tree or Merge Sort: Inversion count algorithms for arbitrary values take $O(N \log N)$ time and $O(N)$ space. Because all values are binary {0, 1}, two-pointer reader-writer simulation solves this in $O(N)$ time and $O(1)$ space.

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

You will see how a senior engineer computes minimum adjacent swaps using reader-writer pointer distances instead of heavy inversion algorithms.

Pattern Recognition Signals

The 10-second spot

Binary digits {0, 1} + adjacent swaps only + move all 0s to one end and 1s to other end.

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

Invariant: arr[0..writer-1] holds compacted target prefix. Next target at reader requires exactly reader - writer adjacent swaps to reach writer.

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

Assuming 0s must always move left: the prompt states 'which end does not matter', requiring evaluation of both zeros-left and ones-left configurations.
Over-engineering with Fenwick trees or Merge Sort: arbitrary inversion counting is O(N log N), but binary arrays permit an exact O(N) two-pointer simulation.
Forgetting to skip leading elements already settled at the destination frontier before beginning reader scan.

The 60-Second Interview Pitch

Say this out loud before you type a single line

"Because only adjacent swaps are allowed, moving a target digit from index reader to index writer across intervening digits takes exactly reader - writer moves. I run two linear passes: one computing swaps to bring all 0s to the left, and one computing swaps to bring all 1s to the left. Returning min(zeros_left, ones_left) runs in O(N) time and O(1) space."

So: test both polarities (zeros left vs ones left) using reader - writer compaction distance in 2N steps and return the minimum.

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(N)

SPACE COMPLEXITY

O(1) Auxiliary

Look at memory allocations: Only integer pointer variables (writer, reader, zero_left_counter, ones_left_counter) are maintained. No auxiliary arrays or recursive stacks are allocated.

Formal Recurrence Relation

$T(N) = O(N)$

Derivation Progression

Zeros-to-Left Compaction Pass

$N \text{ iterations}$

Scans array accumulating reader - writer for all 0s.

Ones-to-Left Compaction Pass

$N \text{ iterations}$

Scans array accumulating reader - writer for all 1s.

Total Operations

$N + N = 2N = O(N)$

Strictly linear time with zero dynamic allocation.

Variable Definitions

$N$

Length of the binary input array

Memory Architecture & Bounds

🟣 Call Stack

O(1) Iterative

🔵 Auxiliary Heap

O(1) No heap allocation

🟢 Output Space

O(1) Scalar integer return

Boundary Best / Worst Cases

Best Case

$O(N)$

Average Case

$O(N)$

Worst Case

$O(N)$

Pointer Invariant Transition Progression

Synthesizing vector architecture diagram...

Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

CONSTRAINTS & BOUNDS

$1 \le n \le 10^5$ , $arr[i] \in \{0, 1\}$ . Requires strictly $O(N)$ time and $O(1)$ auxiliary space.

FAANG PRODUCTION TRAPS & EDGE CASES

Skipping dual-polarity check: the prompt states 'which end does not matter', so evaluating both zeros-to-left and ones-to-left is required to obtain the true minimum.

Core Algorithmic State Invariants

1. Inversion Delta via Reader-Writer Distance

2. Frontier Advancing Compaction Invariant

3. Symmetric Dual-Configuration Minimization

Rosetta Dual-Monaco ComparisonPython 3

CANONICAL INVARIANT TEMPLATE

CONCRETE: MINIMUM MOVES TO SORT BINARY ARRAY (AMAZON OA)

Pattern Implementation Mapping TableCanonical Invariant ⟷ Concrete Code ⟷ Engineering Rationale

Canonical Invariant	Concrete Code	Engineering Rationale
Skip leading target elements already settled at frontier	`while writer0 < n and arr[writer0] == 0: writer0 += 1`	Elements already in their final position require 0 moves; advance destination frontier.
Sequential reader scan over unsorted trailing stretch	`for reader0 in range(writer0, n):`	Reader traverses rightward to discover matching candidate digits.
Inversion delta accumulator for matching target elements	`if arr[reader0] == 0: zero_moves += reader0 - writer0 writer0 += 1`	Bubbling arr[reader0] across intervening elements to writer slot takes exactly (reader0 - writer0) adjacent swaps.
Bilateral polarity evaluation (0s left vs 1s left)	`return min(zero_moves, ones_moves)`	The orientation requirement permits either end; return the minimal move configuration.

Pass

Reader

arr[reader]

Condition / Action

Writer

Delta (reader - writer)

Cumulative Moves

0s Left

Match! Bubble to writer=0

4 - 0 = 4

0s Left

Match! Bubble to writer=1

6 - 1 = 5

9 (Total 0s)

1s Left

Match! Bubble to writer=4

5 - 4 = 1

1s Left

Match! Bubble to writer=5

7 - 5 = 2

3 (Total 1s)

Final

—

Return

\min(9, 3)

—

Pass

Reader

arr[reader]

Condition / Action

Writer

Delta (reader - writer)

Cumulative Moves

0s Left

Match! Bubble to writer=0

4 - 0 = 4

0s Left

Match! Bubble to writer=1

6 - 1 = 5

9 (Total 0s)

1s Left

Match! Bubble to writer=4

5 - 4 = 1

1s Left

Match! Bubble to writer=5

7 - 5 = 2

3 (Total 1s)

Final

—

Return

\min(9, 3)

—

🧭 Conceptual Foundation & Pattern Intuition

🏟️ The Analogy: The Conveyor Belt Sorter

🪄 The Mathematical Harmony / Magic Trick

Code / Blueprint

for reader in range(writer, len(arr)):
    if arr[reader] == target:
        moves += reader - writer
        writer += 1

💡 Summary

Canonical Invariant

Concrete Code

Engineering Rationale

Skip leading target elements already settled at frontier

while writer0 < n and arr[writer0] == 0:
    writer0 += 1

Elements already in their final position require 0 moves; advance destination frontier.

Sequential reader scan over unsorted trailing stretch

for reader0 in range(writer0, n):

Reader traverses rightward to discover matching candidate digits.

Inversion delta accumulator for matching target elements

if arr[reader0] == 0:
    zero_moves += reader0 - writer0
    writer0 += 1

Bubbling arr[reader0] across intervening elements to writer slot takes exactly (reader0 - writer0) adjacent swaps.

Bilateral polarity evaluation (0s left vs 1s left)

return min(zero_moves, ones_moves)

The orientation requirement permits either end; return the minimal move configuration.