Dynamic Programming Series

Day 15

Make Array Strictly Increasing

(Leetcode)

Problem Description

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.

Problem Description

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.

arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]

Problem Description

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.

arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]

arr1 = [1,2,3,6,7]

1 operation

How To approach this problem?

We will first do brute force meaning

try replacing every number from the arr2 check how many operation it will take to make complete array strictly increasing.

	i-1	i	i+1

How To approach this problem?

We will first do brute force meaning

try replacing every number from the arr2 check how many operation it will take to make complete array strictly increasing.

	i-1	i	i+1

arr[i] <= arr[i-1]

arr[i] > arr[i-1]

How To approach this problem?

	i-1	i	i+1

arr[i] <= arr[i-1]

arr[i] > arr[i-1]

It means we have to take an element from arr2 to replace arr1[i]

that element must be greater than the arr[i-1].

Choosing numbers from arr2

arr2 = [a,b,c,d]

a > arr1[i-1]

b > arr1[i-1]

Here we will apply the greedy approach to choose elements from the arr2. Meaning we will choose least greater element.

Choosing numbers from arr2

arr2 = [a,b,c,d]

a > arr1[i-1]

b > arr1[i-1]

Here we will apply the greedy approach to choose elements from the arr2. Meaning we will choose least greater element.

arr1 = 1, 9, 4, 5, 6

arr2 = 2, 8, 12, 13

So we will sort the arr2 and choose the just greater element by applying binary search

How To approach this problem?

	i-1	i	i+1

arr[i] <= arr[i-1]

arr[i] > arr[i-1]

It means we have to take an element from arr2 to replace arr1[i]

that element must be greater than the arr1[i-1].

How To approach this problem?

	i-1	i	i+1

arr[i] <= arr[i-1]

arr[i] > arr[i-1]

Let it be

How To approach this problem?

	i-1	i	i+1

arr[i] <= arr[i-1]

arr[i] > arr[i-1]

Let it be

Replace

How To approach this problem?

	i-1	i	i+1

arr[i] <= arr[i-1]

arr[i] > arr[i-1]

Let it be

Replace

???

How To approach this problem?

	i-1	i	i+1

arr[i] <= arr[i-1]

arr[i] > arr[i-1]

Let it be

Replace

???

arr1 = 1, 2, 100, 5, 6, 7

arr2 = 3, 101, 102, 103

How To approach this problem?

	i-1	i	i+1

arr[i] <= arr[i-1]

arr[i] > arr[i-1]

Let it be

Replace

???

arr1 = 1, 2, 100, 5, 6, 7

arr2 = 3, 101, 102, 103

How To approach this problem?

	i-1	i	i+1

arr[i] <= arr[i-1]

arr[i] > arr[i-1]

Let it be

Replace

Time consuming

How To approach this problem?

We will apply the same approach but with memoization.

Getting the states of memoization

	L	i	i+1

DP[i][L] = min operation required till ith with left of arr1 as value L

Let's try to form the solution of this approach?

arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]

arr1 = 1, 5, 3, 6, 7

arr2 = 1, 2, 3, 4

let arr1[-1] = -inf

hence arr1[0] > arr1[-1]

Let's try to form the solution of this approach?

arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]

arr1 = 1, 5, 3, 6, 7

arr2 = 1, 2, 3, 4

hence arr1[1] > arr1[0]

Let's try to form the solution of this approach?

arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]

arr1 = 1, 5, 3, 6, 7

arr2 = 1, 2, 3, 4

hence arr1[2] < arr1[1]

Not possible configuration

Let's try to form the solution of this approach?

arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]

arr1 = 1, 5, 3, 6, 7

arr2 = 1, 2, 3, 4

hence arr1[2] < arr1[1]

Not possible configuration

arr1 = 1, 2, 3, 6, 7

arr2 = 1, 2, 3, 4

Code

int check(int curr, int left, vector<int>& arr1 ,vector<int>& arr2, vector <unordered_map<int,int> >& dp)
{
    if (curr == arr1.size())
        return 0;
    if (dp[curr].find(left) != dp[curr].end())
        return dp[curr][left];
    int res1,res2;
    if (arr1[curr] > left)    
    {
        res1 = check(curr+1, arr1[curr], arr1, arr2, dp);
    }
    else
    {
        res1 = INT_MAX;
    }
    int rep = upper_bound(arr2.begin(),arr2.end(), left) - arr2.begin();
    if (rep == arr2.size())
        res2 = INT_MAX;
    else
        res2 = check(curr+1, arr2[rep], arr1, arr2, dp);
    if (res2 == INT_MAX)
        dp[curr][left] = res1;
    else
        dp[curr][left] = min(res1,res2+1);
    return dp[curr][left];
}
int makeArrayIncreasing(vector<int>& arr1, vector<int>& arr2) {
    sort(arr2.begin(), arr2.end());
    vector <unordered_map<int,int> > dp(2001);    
    int val = check(0, INT_MIN, arr1, arr2, dp);
    if (val != INT_MAX)
        return val;
    else
        return -1;
}

Time Complexity

Time\ Complexity\ =\ O(n^2 log n)

So that's It for Day 15

Concepts that we have learnt

So that's It for Day 15

We have taken a very first step in Dynamic Programming.

Concepts that we have learnt

1. Dynamic Programming + Greedy

So that's It for Day 15

We have taken a very first step in Dynamic Programming.

Concepts that we have learnt

1. Dynamic Programming + Greedy

2. Use of Brute force with DP.

So that's It for Day 15

We have taken a very first step in Dynamic Programming.

Concepts that we have learnt

1. Dynamic Programming + Greedy

2. Use of Brute force with DP.

Still, Have any Doubts???

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