LeetCode #1187 — HARD

Make Array Strictly Increasing

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

The Problem

Problem Statement

Given two integer arrays arr1 and arr2, return the minimum number of operations (possibly zero) needed to make arr1 strictly increasing.

In one operation, you can choose two indices 0 <= i < arr1.length and 0 <= j < arr2.length and do the assignment arr1[i] = arr2[j].

If there is no way to make arr1 strictly increasing, return -1.

Example 1:

Input: arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]
Output: 1
Explanation: Replace 5 with 2, then arr1 = [1, 2, 3, 6, 7].

Example 2:

Input: arr1 = [1,5,3,6,7], arr2 = [4,3,1]
Output: 2
Explanation: Replace 5 with 3 and then replace 3 with 4. arr1 = [1, 3, 4, 6, 7].

Example 3:

Input: arr1 = [1,5,3,6,7], arr2 = [1,6,3,3]
Output: -1
Explanation: You can't make arr1 strictly increasing.

Constraints:

1 <= arr1.length, arr2.length <= 2000
0 <= arr1[i], arr2[i] <= 10^9

Step 01

Brute Force Baseline

Problem summary: Given two integer arrays arr1 and arr2, return the minimum number of operations (possibly zero) needed to make arr1 strictly increasing. In one operation, you can choose two indices 0 <= i < arr1.length and 0 <= j < arr2.length and do the assignment arr1[i] = arr2[j]. If there is no way to make arr1 strictly increasing, return -1.

Baseline thinking

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

Pattern signal: Array · Binary Search · Dynamic Programming

Example 1

[1,5,3,6,7]
[1,3,2,4]

Example 2

[1,5,3,6,7]
[4,3,1]

Example 3

[1,5,3,6,7]
[1,6,3,3]

Core Insight

What unlocks the optimal approach

Use dynamic programming.
The state would be the index in arr1 and the index of the previous element in arr2 after sorting it and removing duplicates.

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 #1187: Make Array Strictly Increasing
class Solution {
    public int makeArrayIncreasing(int[] arr1, int[] arr2) {
        Arrays.sort(arr2);
        int m = 0;
        for (int x : arr2) {
            if (m == 0 || x != arr2[m - 1]) {
                arr2[m++] = x;
            }
        }
        final int inf = 1 << 30;
        int[] arr = new int[arr1.length + 2];
        arr[0] = -inf;
        arr[arr.length - 1] = inf;
        System.arraycopy(arr1, 0, arr, 1, arr1.length);
        int[] f = new int[arr.length];
        Arrays.fill(f, inf);
        f[0] = 0;
        for (int i = 1; i < arr.length; ++i) {
            if (arr[i - 1] < arr[i]) {
                f[i] = f[i - 1];
            }
            int j = search(arr2, arr[i], m);
            for (int k = 1; k <= Math.min(i - 1, j); ++k) {
                if (arr[i - k - 1] < arr2[j - k]) {
                    f[i] = Math.min(f[i], f[i - k - 1] + k);
                }
            }
        }
        return f[arr.length - 1] >= inf ? -1 : f[arr.length - 1];
    }

    private int search(int[] nums, int x, int n) {
        int l = 0, r = n;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #1187: Make Array Strictly Increasing
func makeArrayIncreasing(arr1 []int, arr2 []int) int {
	sort.Ints(arr2)
	m := 0
	for _, x := range arr2 {
		if m == 0 || x != arr2[m-1] {
			arr2[m] = x
			m++
		}
	}
	arr2 = arr2[:m]
	const inf = 1 << 30
	arr1 = append([]int{-inf}, arr1...)
	arr1 = append(arr1, inf)
	n := len(arr1)
	f := make([]int, n)
	for i := range f {
		f[i] = inf
	}
	f[0] = 0
	for i := 1; i < n; i++ {
		if arr1[i-1] < arr1[i] {
			f[i] = f[i-1]
		}
		j := sort.SearchInts(arr2, arr1[i])
		for k := 1; k <= min(i-1, j); k++ {
			if arr1[i-k-1] < arr2[j-k] {
				f[i] = min(f[i], f[i-k-1]+k)
			}
		}
	}
	if f[n-1] >= inf {
		return -1
	}
	return f[n-1]
}

# Accepted solution for LeetCode #1187: Make Array Strictly Increasing
class Solution:
    def makeArrayIncreasing(self, arr1: List[int], arr2: List[int]) -> int:
        arr2.sort()
        m = 0
        for x in arr2:
            if m == 0 or x != arr2[m - 1]:
                arr2[m] = x
                m += 1
        arr2 = arr2[:m]
        arr = [-inf] + arr1 + [inf]
        n = len(arr)
        f = [inf] * n
        f[0] = 0
        for i in range(1, n):
            if arr[i - 1] < arr[i]:
                f[i] = f[i - 1]
            j = bisect_left(arr2, arr[i])
            for k in range(1, min(i - 1, j) + 1):
                if arr[i - k - 1] < arr2[j - k]:
                    f[i] = min(f[i], f[i - k - 1] + k)
        return -1 if f[n - 1] >= inf else f[n - 1]

// Accepted solution for LeetCode #1187: Make Array Strictly Increasing
/**
 * [1187] Make Array Strictly Increasing
 *
 * Given two integer arrays arr1 and arr2, return the minimum number of operations (possibly zero) needed to make arr1 strictly increasing.
 * In one operation, you can choose two indices 0 <= i < arr1.length and 0 <= j < arr2.length and do the assignment arr1[i] = arr2[j].
 * If there is no way to make arr1 strictly increasing, return -1.
 *  
 * Example 1:
 *
 * Input: arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]
 * Output: 1
 * Explanation: Replace 5 with 2, then arr1 = [1, 2, 3, 6, 7].
 *
 * Example 2:
 *
 * Input: arr1 = [1,5,3,6,7], arr2 = [4,3,1]
 * Output: 2
 * Explanation: Replace 5 with 3 and then replace 3 with 4. arr1 = [1, 3, 4, 6, 7].
 *
 * Example 3:
 *
 * Input: arr1 = [1,5,3,6,7], arr2 = [1,6,3,3]
 * Output: -1
 * Explanation: You can't make arr1 strictly increasing.
 *  
 * Constraints:
 *
 * 	1 <= arr1.length, arr2.length <= 2000
 * 	0 <= arr1[i], arr2[i] <= 10^9
 *
 *  
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/make-array-strictly-increasing/
// discuss: https://leetcode.com/problems/make-array-strictly-increasing/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    // Credit: https://leetcode.com/problems/make-array-strictly-increasing/solutions/3647105/rust-dp-solution/
    pub fn make_array_increasing(arr1: Vec<i32>, arr2: Vec<i32>) -> i32 {
        let mut arr2 = arr2;
        arr2.sort();
        arr2.dedup();

        let inf = 1 << 30;
        let mut arr1 = arr1;
        arr1.push(inf);
        arr1.insert(0, -inf);

        let mut dp = vec![inf; arr1.len()];
        dp[0] = 0;

        for i in 1..arr1.len() {
            if arr1[i - 1] < arr1[i] {
                dp[i] = dp[i - 1];
            }

            let j = match arr2.binary_search(&arr1[i]) {
                Ok(pos) => pos,
                Err(pos) => pos,
            };

            for k in 1..=j.min(i - 1) {
                if arr1[i - k - 1] < arr2[j - k] {
                    dp[i] = dp[i].min(dp[i - k - 1] + k as i32);
                }
            }
        }

        if dp[arr1.len() - 1] >= inf {
            -1
        } else {
            dp[arr1.len() - 1]
        }
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1187_example_1() {
        let arr1 = vec![1, 5, 3, 6, 7];
        let arr2 = vec![1, 3, 2, 4];
        let result = 1;

        assert_eq!(Solution::make_array_increasing(arr1, arr2), result);
    }

    #[test]
    fn test_1187_example_2() {
        let arr1 = vec![1, 5, 3, 6, 7];
        let arr2 = vec![4, 3, 1];
        let result = 2;

        assert_eq!(Solution::make_array_increasing(arr1, arr2), result);
    }

    #[test]
    fn test_1187_example_3() {
        let arr1 = vec![1, 5, 3, 6, 7];
        let arr2 = vec![1, 6, 3, 3];
        let result = -1;

        assert_eq!(Solution::make_array_increasing(arr1, arr2), result);
    }
}

// Accepted solution for LeetCode #1187: Make Array Strictly Increasing
function makeArrayIncreasing(arr1: number[], arr2: number[]): number {
    arr2.sort((a, b) => a - b);
    let m = 0;
    for (const x of arr2) {
        if (m === 0 || x !== arr2[m - 1]) {
            arr2[m++] = x;
        }
    }
    arr2 = arr2.slice(0, m);
    const inf = 1 << 30;
    arr1 = [-inf, ...arr1, inf];
    const n = arr1.length;
    const f: number[] = new Array(n).fill(inf);
    f[0] = 0;
    const search = (arr: number[], x: number): number => {
        let l = 0;
        let r = arr.length;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (arr[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    for (let i = 1; i < n; ++i) {
        if (arr1[i - 1] < arr1[i]) {
            f[i] = f[i - 1];
        }
        const j = search(arr2, arr1[i]);
        for (let k = 1; k <= Math.min(i - 1, j); ++k) {
            if (arr1[i - k - 1] < arr2[j - k]) {
                f[i] = Math.min(f[i], f[i - k - 1] + k);
            }
        }
    }
    return f[n - 1] >= inf ? -1 : f[n - 1];
}

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(log n)

Space

O(1)

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.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.