LeetCode #1713 — HARD

Minimum Operations to Make a Subsequence

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an array target that consists of distinct integers and another integer array arr that can have duplicates.

In one operation, you can insert any integer at any position in arr. For example, if arr = [1,4,1,2], you can add 3 in the middle and make it [1,4,3,1,2]. Note that you can insert the integer at the very beginning or end of the array.

Return the minimum number of operations needed to make target a subsequence of arr.

A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements' relative order. For example, [2,7,4] is a subsequence of [4,2,3,7,2,1,4] (the underlined elements), while [2,4,2] is not.

Example 1:

Input: target = [5,1,3], arr = [9,4,2,3,4]
Output: 2
Explanation: You can add 5 and 1 in such a way that makes arr = [5,9,4,1,2,3,4], then target will be a subsequence of arr.

Example 2:

Input: target = [6,4,8,1,3,2], arr = [4,7,6,2,3,8,6,1]
Output: 3

Constraints:

1 <= target.length, arr.length <= 10⁵
1 <= target[i], arr[i] <= 10⁹
target contains no duplicates.

Step 01

Brute Force Baseline

Problem summary: You are given an array target that consists of distinct integers and another integer array arr that can have duplicates. In one operation, you can insert any integer at any position in arr. For example, if arr = [1,4,1,2], you can add 3 in the middle and make it [1,4,3,1,2]. Note that you can insert the integer at the very beginning or end of the array. Return the minimum number of operations needed to make target a subsequence of arr. A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements' relative order. For example, [2,7,4] is a subsequence of [4,2,3,7,2,1,4] (the underlined elements), while [2,4,2] is not.

Baseline thinking

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

Pattern signal: Array · Hash Map · Binary Search · Greedy

Example 1

[5,1,3]
[9,4,2,3,4]

Example 2

[6,4,8,1,3,2]
[4,7,6,2,3,8,6,1]

Core Insight

What unlocks the optimal approach

The problem can be reduced to computing Longest Common Subsequence between both arrays.
Since one of the arrays has distinct elements, we can consider that these elements describe an arrangement of numbers, and we can replace each element in the other array with the index it appeared at in the first array.
Then the problem is converted to finding Longest Increasing Subsequence in the second array, which can be done in O(n log n).

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

Largest constraint values

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 #1713: Minimum Operations to Make a Subsequence
class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        this.c = new int[n + 1];
    }

    public void update(int x, int v) {
        for (; x <= n; x += x & -x) {
            c[x] = Math.max(c[x], v);
        }
    }

    public int query(int x) {
        int ans = 0;
        for (; x > 0; x -= x & -x) {
            ans = Math.max(ans, c[x]);
        }
        return ans;
    }
}

class Solution {
    public int minOperations(int[] target, int[] arr) {
        int m = target.length;
        Map<Integer, Integer> d = new HashMap<>(m);
        for (int i = 0; i < m; i++) {
            d.put(target[i], i + 1);
        }
        List<Integer> nums = new ArrayList<>();
        for (int x : arr) {
            if (d.containsKey(x)) {
                nums.add(d.get(x));
            }
        }
        BinaryIndexedTree tree = new BinaryIndexedTree(m);
        int ans = 0;
        for (int x : nums) {
            int v = tree.query(x - 1) + 1;
            ans = Math.max(ans, v);
            tree.update(x, v);
        }
        return m - ans;
    }
}

// Accepted solution for LeetCode #1713: Minimum Operations to Make a Subsequence
type BinaryIndexedTree struct {
	n int
	c []int
}

func NewBinaryIndexedTree(n int) BinaryIndexedTree {
	return BinaryIndexedTree{n: n, c: make([]int, n+1)}
}

func (bit *BinaryIndexedTree) Update(x, v int) {
	for ; x <= bit.n; x += x & -x {
		if v > bit.c[x] {
			bit.c[x] = v
		}
	}
}

func (bit *BinaryIndexedTree) Query(x int) int {
	ans := 0
	for ; x > 0; x -= x & -x {
		if bit.c[x] > ans {
			ans = bit.c[x]
		}
	}
	return ans
}

func minOperations(target []int, arr []int) int {
	m := len(target)
	d := make(map[int]int)
	for i, x := range target {
		d[x] = i + 1
	}
	var nums []int
	for _, x := range arr {
		if pos, exists := d[x]; exists {
			nums = append(nums, pos)
		}
	}
	tree := NewBinaryIndexedTree(m)
	ans := 0
	for _, x := range nums {
		v := tree.Query(x-1) + 1
		if v > ans {
			ans = v
		}
		tree.Update(x, v)
	}
	return m - ans
}

# Accepted solution for LeetCode #1713: Minimum Operations to Make a Subsequence
class BinaryIndexedTree:
    __slots__ = "n", "c"

    def __init__(self, n: int):
        self.n = n
        self.c = [0] * (n + 1)

    def update(self, x: int, v: int):
        while x <= self.n:
            self.c[x] = max(self.c[x], v)
            x += x & -x

    def query(self, x: int) -> int:
        res = 0
        while x:
            res = max(res, self.c[x])
            x -= x & -x
        return res


class Solution:
    def minOperations(self, target: List[int], arr: List[int]) -> int:
        d = {x: i for i, x in enumerate(target, 1)}
        nums = [d[x] for x in arr if x in d]
        m = len(target)
        tree = BinaryIndexedTree(m)
        ans = 0
        for x in nums:
            v = tree.query(x - 1) + 1
            ans = max(ans, v)
            tree.update(x, v)
        return len(target) - ans

// Accepted solution for LeetCode #1713: Minimum Operations to Make a Subsequence
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn min_operations(target: Vec<i32>, arr: Vec<i32>) -> i32 {
        let n = target.len();
        let mut idx: HashMap<i32, usize> = HashMap::new();
        for x in target {
            idx.insert(x, idx.len());
        }
        let mut dp = vec![];
        for x in arr {
            if let Some(&i) = idx.get(&x) {
                let j = match dp.binary_search(&i) {
                    Ok(j) => j,
                    Err(j) => j,
                };
                if j < dp.len() {
                    dp[j] = i;
                } else {
                    dp.push(i);
                }
            }
        }
        (n - dp.len()) as i32
    }
}

#[test]
fn test() {
    let target = vec![5, 1, 3];
    let arr = vec![9, 4, 2, 3, 4];
    let res = 2;
    assert_eq!(Solution::min_operations(target, arr), res);
    let target = vec![6, 4, 8, 1, 3, 2];
    let arr = vec![4, 7, 6, 2, 3, 8, 6, 1];
    let res = 3;
    assert_eq!(Solution::min_operations(target, arr), res);
}

// Accepted solution for LeetCode #1713: Minimum Operations to Make a Subsequence
class BinaryIndexedTree {
    private n: number;
    private c: number[];

    constructor(n: number) {
        this.n = n;
        this.c = Array(n + 1).fill(0);
    }

    update(x: number, v: number): void {
        for (; x <= this.n; x += x & -x) {
            this.c[x] = Math.max(this.c[x], v);
        }
    }

    query(x: number): number {
        let ans = 0;
        for (; x > 0; x -= x & -x) {
            ans = Math.max(ans, this.c[x]);
        }
        return ans;
    }
}

function minOperations(target: number[], arr: number[]): number {
    const m = target.length;
    const d: Map<number, number> = new Map();
    target.forEach((x, i) => d.set(x, i + 1));
    const nums: number[] = [];
    arr.forEach(x => {
        if (d.has(x)) {
            nums.push(d.get(x)!);
        }
    });
    const tree = new BinaryIndexedTree(m);
    let ans = 0;
    nums.forEach(x => {
        const v = tree.query(x - 1) + 1;
        ans = Math.max(ans, v);
        tree.update(x, v);
    });
    return m - 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 m)

Space

O(m)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

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.

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.