**Infosys hackwithinfy Sample coding questions |** **Infosys hackwithinfy coding questions 2024** **| Hackwithinfy coding questions 2024:** Are you gearing up for the Infosys 2024 HackWithInfy registration? If you’re looking to participate and prepare for the upcoming Infosys HackWithInfy hackathon, you’ve landed on the perfect platform. Explore Hackwithinfy coding questions 2024 or previous year’s HackWithInfy coding questions and gear up for the challenge ahead. Get ready to ace the Infosys HackWithInfy coding competition!

Below are sample questions provided by Infosys company itself. These will help you understand the difficulty level of HackWithInfy coding questions in the exam.

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**Infosys Hackwithinfy Sample coding questions:**

**Problem Statement 1:** **of Infosys Hackwithinfy**

**Today you decided to go to the gym. You currently have E energy. There are N exercises in the gym. Each of these exercises drains Ai amount of energy from your body. You feel tired if your energy reaches 0 or below. Calculate the minimum number of exercises you have to perform such that you become tired. Every unique exercise can only be performed at most 2 times as others also have to use the machines. If performing all the exercises does not make you feel tired, return -1.**

**Parameters:**

E :: INTEGER

The first line contains an integer, E, denoting the Energy. E :: 1 -> 10^5

N :: INTEGER

The next line contains an integer, N, denoting the number of exercises.

N :: 1 -> 10^5

A :: INTEGER ARRAY

Each line i of the N subsequent lines (where 0 ≤ i < N) contains an integer describing the amount of energy drained by ith exercise. A[i] :: 1 -> 10^5

Case#: 1**Input:**

6

2

1

2**Output:**

4

E = 6

Do 1st exercise 2 times Do 2nd exercise 2 times

Hence, total exercise done 4. Case#: 2**Input:**

10

2

1

2

**Output:**

-1

E = 10

By doing both the exercises 2 times you won’t feel tired. Case#: 3

**Input:**

2

3

1

5

2**Output:**

1

E = 2

Use 3rd exercise 1 time. Hence, total exercise done 1.

**Problem Statement 2:** **of Infosys Hackwithinfy**

**There is a battle between heroes and villains going on. You have M heroes, all of them have the same health H. There are N villains, health of the ith villain is Vi.**

When a hero, with health H battles a villain with health Vi, one of the three scenarios can happen:

if H > Vi: The villain is defeated and the health of the hero is decreased by Vi

if H < Vi: The villain wins, his health is not affected and the hero is no longer able to fight. if H = Vi: Both of them are considered defeated and neither can fight. The heroes start fighting villains one by one in the same order, first villain 1 then villain 2 and so on. It is might be possible that before defeating all the villains, all the heroes are defeated. Therefore, to ensure the victory of the heroes, you want to remove some villains from the front. Your task is to find the minimum number of villains you need to remove from the front such that the victory of the heroes is guaranteed. Note: If in the last battle, both the hero and villain are defeated and no more heroes or villains remain, it would still be considered a victory since all the villains are defeated. Parameters: N :: INTEGER The first line contains an integer, N, denoting the number of villains N :: 1 -> 2*10^5 M :: INTEGER The next line contains an integer, M, denoting the number of heroes M :: 1 -> 2*10^5 H :: INTEGER

The next line contains an integer, H, denoting the health of each of the heroes

H :: 1 -> 10^9

array :: INTEGER ARRAY

Each line i of the N subsequent lines (where 0 ≤ i < N) contains an integer describing the health of each of the villains. array[i] :: 1 -> 10^9 Case#: 1

Input:

4

4

3

3

1

3

3**Output:**

0

[3, 1, 3, 3]. We have 4 heroes will health 3. The heroes 1 will fight villain 1. Both get defeated. The hero 2 fights villain 2. It wins the battle and now his health is 2. He fights the third villain and loses, the villain still has health 3. The hero 3 fights villain 3 and both get defeated. Hero 4 fights villain 4 and both get defeated. So no need to remove any villain.

Case#: 2**Input:**

5

3

3

1

2

3

1

1**Output:**

0

The fight will take place and hero 1 will defeat villain 1 and 2. Hero 2 will defeat villain 2. Hero 3 will defeat villain 3 and 4

Case#: 3**Input:**

5

1

4

1

2

3

1

3**Output:**

3

Only 1 hero is present with health 4. Since you can only remove villain from the front, you will have to remove the first 3 villains to ensure victory. The hero can fight the last 2 villain of health 1 and 3 respectively and win the battle.

**Problem Statement 2:** **of Infosys Hackwithinfy**

**You need to build a road in a rugged terrain. You know the sea level of each segment of the rugged terrain, i.e. the i-th segment is Li meters from sea level.You needs to transform the terrain into a strictly downward sloping terrain for the road, i.e, for each i-th segment where 2 <= i <= N, resultant Li-1 > Li. In order to do so, you employ a powerful digging team to help you dig and reduce the sea level of the segments. On day D, the team can reduce the sea level for each segments that you scheduled that day by 2D-1 meters each.You are allowed to assign the team to dig on multiple segments and/or dig on the same segments for multiple days.Your task is to find the minimum number of days needed to transform the terrain as per your requirements.**

Parameters:

N :: INTEGER

The first line contains an integer, N, denoting the number of elements in L.

N :: 1 -> 10^5

L :: INTEGER ARRAY

Each line i of the N subsequent lines (where 0 < i ≤ N) contains an integer describing Li, the sea level of the i-th segment. L[i] :: -10^9 -> 10^9

Case#: 1

**Input:**

2

3

3

**Output:**

1

We can dig on the 2nd segment, reducing it from 3-meter sea level to 2. Resulting in {3, 2} which is strictly decreasing.

**Case#: 2 Input:**

2

5

-3

**Output:**

0

It is already strictly decreasing before start. Case#: 3

Input:

4

-1

1

1

1

Output:

3

One of the possible way:

On day 1, we can dig on 1st and 4th segment, resulting in {-2, 1, 1, 0}

On day 2, we can dig on 3rd and 4th segments, resulting in {-2, 1,

-1, -2}

On day 3, we can dig on 2nd, 3rd and 4th segments, resulting in

{-2, -3, -5, -6}

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