Maximum score possible from an array with jumps of at most length K
Last Updated :
28 Feb, 2025
Given an array arr[] and an integer k. The task is to find the maximum score we can achieve by performing the following operations:
- Start at the 0th index of the array.
- Move to the last index of the array by jumping at most k indices at a time. For example, from the index i, we can jump to any index between i + 1 and i + k (inclusive) as long as it is within the bounds of the array.
- Collect the value of each index we land on, including the value at the starting index (0th index).
Examples:
Input: arr[] = [100, -30, -50, -15, -20, -30], k = 3
Output: 55
Explanation: From 0th index, jump 3 indices ahead to arr[3]. From 3rd, jump 2 steps ahead to arr[5]. Therefore, the maximum score possible = (100 + (-15) + (-30)) = 55
Input: arr[] = [-44, -17, -54, 79], k = 2
Output: 18
Explanation: From 0th index, jump 1 index ahead to arr[1]. From index 1, jump 2 steps ahead to arr[3]. Therefore, the maximum score possible = -44 + (-17) + 79 = 18.
Using Top-Down DP (Memoization) - O(n*k) Time and O(n) Space
The idea is to use recursion with memoization to explore all possible paths from the first index to the last, aiming to find the maximum score.
- Starting from any index, the function calculates the score by adding the current index value and the score obtained from valid jumps within a range of k.
- If the last index is reached, the value at that index is returned as the base case.
- If a result for the current index has already been computed, it is retrieved from the memo[] array to avoid redundant calculations.
Steps to implement the above idea:
- Initialize a memoization array with -1 to store computed results.
- Define a recursive function that computes the maximum score from the current index, checking memoized values to avoid recomputation.
- Base case: If at the last index, return its value.
- Iterate through jumps (from 1 to k), recursively computing the score for each valid jump.
- Memoize the maximum score obtained from all possible jumps at the current index.
C++
// C++ implamentation to find max score
// from at most k jumps using
// Recursion + Memoization
#include <bits/stdc++.h>
using namespace std;
// Function to calculate maximum score
int scoreHelper(vector<int> &arr, int k,
int index, vector<int> &memo) {
// Return memoized value if already computed
if (memo[index] != -1) {
return memo[index];
}
// Return value at the last index if reached
if (index == arr.size() - 1) {
return arr[index];
}
// Initialize maxScore for current index
int maxScore = INT_MIN;
// Explore all valid jumps within range
for (int jump = 1; jump <= k; jump++) {
if (index + jump < arr.size()) {
maxScore = max(maxScore, arr[index] +
scoreHelper(arr, k, index + jump, memo));
}
}
// Memoize the result for current index
memo[index] = maxScore;
return memo[index];
}
// Main function to get maximum score
int getScore(vector<int> &arr, int k) {
// Memoization array
vector<int> memo(arr.size(), -1);
return scoreHelper(arr, k, 0, memo);
}
int main() {
vector<int> arr = {100, -30, -50, -15, -20, -30};
int k = 3;
cout << getScore(arr, k) << endl;
return 0;
}
// Java implementation to find max score
// from at most k jumps using
// Recursion + Memoization
import java.util.*;
class GfG {
// Function to calculate maximum score
static int scoreHelper(int[] arr, int k,
int index, int[] memo) {
// Return memoized value if already computed
if (memo[index] != -1) {
return memo[index];
}
// Return value at the last index if reached
if (index == arr.length - 1) {
return arr[index];
}
// Initialize maxScore for current index
int maxScore = Integer.MIN_VALUE;
// Explore all valid jumps within range
for (int jump = 1; jump <= k; jump++) {
if (index + jump < arr.length) {
maxScore = Math.max(maxScore,
arr[index] + scoreHelper(arr, k,
index + jump, memo));
}
}
// Memoize the result for current index
memo[index] = maxScore;
return memo[index];
}
// Main function to get maximum score
static int getScore(int[] arr, int k) {
// Memoization array
int[] memo = new int[arr.length];
Arrays.fill(memo, -1);
return scoreHelper(arr, k, 0, memo);
}
public static void main(String[] args) {
int[] arr = {100, -30, -50, -15, -20, -30};
int k = 3;
System.out.println(getScore(arr, k));
}
}
# Python implementation to find max score
# from at most k jumps using
# Recursion + Memoization
# Function to calculate maximum score
def scoreHelper(arr, k, index, memo):
# Return memoized value if already computed
if memo[index] != -1:
return memo[index]
# Return value at the last index if reached
if index == len(arr) - 1:
return arr[index]
# Initialize maxScore for current index
maxScore = float('-inf')
# Explore all valid jumps within range
for jump in range(1, k + 1):
if index + jump < len(arr):
maxScore = max(
maxScore,
arr[index] + scoreHelper(arr, k, index + jump, memo)
)
# Memoize the result for current index
memo[index] = maxScore
return memo[index]
# Main function to get maximum score
def getScore(arr, k):
# Memoization array
memo = [-1] * len(arr)
return scoreHelper(arr, k, 0, memo)
if __name__ == "__main__":
arr = [100, -30, -50, -15, -20, -30]
k = 3
print(getScore(arr, k))
// C# implementation to find max score
// from at most k jumps using
// Recursion + Memoization
using System;
class GfG {
// Function to calculate maximum score
static int ScoreHelper(int[] arr, int k,
int index, int[] memo) {
// Return memoized value if already computed
if (memo[index] != -1) {
return memo[index];
}
// Return value at the last index if reached
if (index == arr.Length - 1) {
return arr[index];
}
// Initialize maxScore for current index
int maxScore = int.MinValue;
// Explore all valid jumps within range
for (int jump = 1; jump <= k; jump++) {
if (index + jump < arr.Length) {
maxScore = Math.Max(
maxScore,
arr[index]
+ ScoreHelper(arr, k, index + jump,
memo));
}
}
// Memoize the result for current index
memo[index] = maxScore;
return memo[index];
}
// Main function to get maximum score
static int GetScore(int[] arr, int k) {
// Memoization array
int[] memo = new int[arr.Length];
Array.Fill(memo, -1);
return ScoreHelper(arr, k, 0, memo);
}
static void Main(string[] args) {
int[] arr = { 100, -30, -50, -15, -20, -30 };
int k = 3;
Console.WriteLine(GetScore(arr, k));
}
}
// JavaScript implementation to find max score
// from at most k jumps using
// Recursion + Memoization
// Function to calculate maximum score
function scoreHelper(arr, k, index, memo) {
// Return memoized value if already computed
if (memo[index] !== -1) {
return memo[index];
}
// Return value at the last index if reached
if (index === arr.length - 1) {
return arr[index];
}
// Initialize maxScore for current index
let maxScore = -Infinity;
// Explore all valid jumps within range
for (let jump = 1; jump <= k; jump++) {
if (index + jump < arr.length) {
maxScore = Math.max(
maxScore,
arr[index]
+ scoreHelper(arr, k, index + jump,
memo));
}
}
// Memoize the result for current index
memo[index] = maxScore;
return memo[index];
}
// Main function to get maximum score
function getScore(arr, k) {
// Memoization array
let memo = new Array(arr.length).fill(-1);
return scoreHelper(arr, k, 0, memo);
}
let arr = [ 100, -30, -50, -15, -20, -30 ];
let k = 3;
console.log(getScore(arr, k));
Using Bottom-Up DP (Tabulation) - O(n*k) Time and O(n) Space
The idea is to use dynamic programming with a bottom-up approach to find the maximum score:
- Instead of recursively exploring paths, a dp array stores the best possible score from each index to the end.
- The solution builds backward, ensuring each index considers the best jump within k steps using previously computed results.
- This avoids redundant calculations, making it more efficient than recursion. The final answer is found at dp[0], representing the maximum score from the start.
Steps to implement the above idea:
- Initialize a dp array of size n with minimum integer value.
- Set the last index of dp as arr[n-1] (base case).
- Iterate backward from the second last index to the first.
- For each index, check all valid jumps up to k steps.
- Update dp[i] as the maximum possible score from that index.
- Return dp[0], which gives the maximum score from the start.
C++
// C++ implementation to find max score
// from at most k jumps using Tabulation
#include <bits/stdc++.h>
using namespace std;
// Function to calculate maximum score using Tabulation
int getScore(vector<int> &arr, int k) {
int n = arr.size();
// Tabulation array to store maximum scores
vector<int> dp(n, INT_MIN);
// Base case: score at the last index is the
// value at that index
dp[n - 1] = arr[n - 1];
// Iterate from second last index to the first
for (int i = n - 2; i >= 0; i--) {
// Calculate max score by considering all valid jumps
for (int jump = 1; jump <= k; jump++) {
if (i + jump < n) {
dp[i] = max(dp[i], arr[i] + dp[i + jump]);
}
}
}
// Return the maximum score starting
// from the first index
return dp[0];
}
int main() {
vector<int> arr = {100, -30, -50, -15, -20, -30};
int k = 3;
cout << getScore(arr, k) << endl;
return 0;
}
// Java implementation to find max score
// from at most k jumps using Tabulation
import java.util.*;
class GfG {
// Main function to get maximum score
static int getScore(int[] arr, int k) {
int n = arr.length;
// Tabulation array to store scores
int[] dp = new int[n];
// Base case: score at the last index
dp[n - 1] = arr[n - 1];
// Iterate from second last index to first
for (int i = n - 2; i >= 0; i--) {
// Initialize maxScore for current index
dp[i] = Integer.MIN_VALUE;
// Calculate max score for valid jumps
for (int jump = 1; jump <= k; jump++) {
if (i + jump < n) {
dp[i] = Math.max(dp[i],
arr[i] + dp[i + jump]);
}
}
}
// Return score from the first index
return dp[0];
}
public static void main(String[] args) {
int[] arr = { 100, -30, -50, -15, -20, -30 };
int k = 3;
System.out.println(getScore(arr, k));
}
}
# Python implementation to find max score
# from at most k jumps using Tabulation
# Main function to get maximum score
def getScore(arr, k):
n = len(arr)
# Tabulation array to store scores
dp = [float('-inf')] * n
# Base case: score at the last index
dp[n - 1] = arr[n - 1]
# Iterate from second last index to first
for i in range(n - 2, -1, -1):
# Calculate max score for all valid jumps
for jump in range(1, k + 1):
if i + jump < n:
dp[i] = max(dp[i],
arr[i] + dp[i + jump])
# Return score from the first index
return dp[0]
if __name__ == "__main__":
arr = [100, -30, -50, -15, -20, -30]
k = 3
print(getScore(arr, k))
// C# implementation to find max score
// from at most k jumps using Tabulation
using System;
class GfG {
// Main function to get maximum score
static int GetScore(int[] arr, int k) {
int n = arr.Length;
// Tabulation array to store scores
int[] dp = new int[n];
Array.Fill(dp, int.MinValue);
// Base case: score at the last index
dp[n - 1] = arr[n - 1];
// Iterate from second last index to first
for (int i = n - 2; i >= 0; i--) {
// Calculate max score for all valid jumps
for (int jump = 1; jump <= k; jump++) {
if (i + jump < n) {
dp[i] = Math.Max(dp[i],
arr[i] + dp[i + jump]);
}
}
}
// Return score from the first index
return dp[0];
}
static void Main(string[] args) {
int[] arr = { 100, -30, -50, -15, -20, -30 };
int k = 3;
Console.WriteLine(GetScore(arr, k));
}
}
// JavaScript implementation to find max score
// from at most k jumps using Tabulation
// Main function to get maximum score
function getScore(arr, k) {
let n = arr.length;
// Tabulation array to store scores
let dp = new Array(n).fill(-Infinity);
// Base case: score at the last index
dp[n - 1] = arr[n - 1];
// Iterate from second last index to first
for (let i = n - 2; i >= 0; i--) {
// Calculate max score for all valid jumps
for (let jump = 1; jump <= k; jump++) {
if (i + jump < n) {
dp[i] = Math.max(dp[i],
arr[i] + dp[i + jump]);
}
}
}
// Return score from the first index
return dp[0];
}
let arr = [ 100, -30, -50, -15, -20, -30 ];
let k = 3;
console.log(getScore(arr, k));
Using DP + Heap - O(n*log(k)) Time and O(k) Space
The idea is to improve the efficiency of the basic tabulation method by using a max-heap.
- In the simple tabulation method, for each index, all possible jumps within the range k are evaluated, which can be time-consuming when k is large.
- To optimize this, the max-heap dynamically tracks the maximum scores within the valid range, ensuring faster access to the best possible jump.
- The heap allows the algorithm to focus only on the relevant scores, making the computation significantly faster.
Steps to implement the above idea:
- Initialize a dp array of size n with minimum integer value.
- Set dp[n-1] as arr[n-1] and push it into a max-heap.
- Iterate backward from the second last index to the first.
- Remove out-of-range elements from the heap (index > i + k).
- Set dp[i] as the max heap's top value + arr[i].
- Push dp[i] and its index into the max-heap and return dp[0].
C++
// C++ implementation to find max score
// from at most k jumps using Tabulation + Heap
#include <bits/stdc++.h>
using namespace std;
int getScore(vector<int> &arr, int k) {
int n = arr.size();
vector<int> dp(n, INT_MIN);
// Base case: score at the last index is the
// value at that index
dp[n - 1] = arr[n - 1];
// Max-heap to store {dp[index], index}
priority_queue<pair<int, int>> maxh;
maxh.push({dp[n - 1], n - 1});
// Iterate from second last index to the first
for (int i = n - 2; i >= 0; i--) {
// Remove out-of-range elements from the heap
while (maxh.size() && maxh.top().second > i + k) {
maxh.pop();
}
// Set dp[i] as the max value from the valid
// range + current index value
dp[i] = maxh.top().first + arr[i];
// Push current dp[i] and index into the heap
maxh.push({dp[i], i});
}
// Return the maximum score starting from the
// first index
return dp[0];
}
int main() {
vector<int> arr = {100, -30, -50, -15, -20, -30};
int k = 3;
cout << getScore(arr, k) << endl;
return 0;
}
// Java implementation to find max score
// from at most k jumps using Tabulation + Heap
import java.util.*;
class GfG {
static int getScore(int[] arr, int k) {
int n = arr.length;
// Tabulation array to store scores
int[] dp = new int[n];
// Base case: score at the last index
dp[n - 1] = arr[n - 1];
// Max-heap to store {dp[index], index}
PriorityQueue<int[]> maxh = new PriorityQueue<>(
(a, b) -> Integer.compare(b[0], a[0]));
maxh.offer(new int[] { dp[n - 1], n - 1 });
// Iterate from second last index to first
for (int i = n - 2; i >= 0; i--) {
// Remove out-of-range elements from the heap
while (!maxh.isEmpty()
&& maxh.peek()[1] > i + k) {
maxh.poll();
}
// Set dp[i] as the max value from the valid
// range
// + current index value
dp[i] = maxh.peek()[0] + arr[i];
// Push current dp[i] and index into the heap
maxh.offer(new int[] { dp[i], i });
}
// Return score from the first index
return dp[0];
}
public static void main(String[] args) {
int[] arr = { 100, -30, -50, -15, -20, -30 };
int k = 3;
System.out.println(getScore(arr, k));
}
}
# Python implementation to find max score
# from at most k jumps using Tabulation + Heap
import heapq
def getScore(arr, k):
n = len(arr)
# Tabulation array to store maximum scores
dp = [float('-inf')] * n
# Base case: score at the last index is the
# value at that index
dp[n - 1] = arr[n - 1]
# Max-heap to store {dp[index], index}
maxh = []
heapq.heappush(maxh, (-dp[n - 1], n - 1))
# Iterate from second last index to the first
for i in range(n - 2, -1, -1):
# Remove out-of-range elements from the heap
while maxh and maxh[0][1] > i + k:
heapq.heappop(maxh)
# Set dp[i] as the max value from the valid
# range + current index value
dp[i] = -maxh[0][0] + arr[i]
# Push current dp[i] and index into the heap
heapq.heappush(maxh, (-dp[i], i))
# Return the maximum score starting from the
# first index
return dp[0]
if __name__ == "__main__":
arr = [100, -30, -50, -15, -20, -30]
k = 3
print(getScore(arr, k))
// C# implementation to find max score
// from at most k jumps using Tabulation + Heap
using System;
using System.Collections.Generic;
class GfG {
static int GetScore(int[] arr, int k) {
int n = arr.Length;
// Tabulation array to store scores
int[] dp = new int[n];
// Base case: score at the last index
dp[n - 1] = arr[n - 1];
// Max Heap to store indices of dp values
// in decreasing order
LinkedList<int> maxh = new LinkedList<int>();
maxh.AddLast(n - 1);
// Iterate from second last index to the first
for (int i = n - 2; i >= 0; i--) {
// Remove indices from the heap that are out of range
while (maxh.Count > 0 && maxh.First.Value > i + k) {
maxh.RemoveFirst();
}
// Set dp[i] as the maximum score from the heap
dp[i] = arr[i] + dp[maxh.First.Value];
// Maintain the heap in decreasing order of dp values
while (maxh.Count > 0 && dp[i] >= dp[maxh.Last.Value]) {
maxh.RemoveLast();
}
// Add the current index to the heap
maxh.AddLast(i);
}
// Return the maximum score starting
// from the first index
return dp[0];
}
static void Main(string[] args) {
int[] arr = { 100, -30, -50, -15, -20, -30 };
int k = 3;
Console.WriteLine(GetScore(arr, k));
}
}
// JavaScript implementation to find max score
// from at most k jumps using Tabulation + Heap
function getScore(arr, k) {
let n = arr.length;
// Tabulation array to store maximum scores
let dp = new Array(n).fill(Number.NEGATIVE_INFINITY);
// Base case: score at the last index is the
// value at that index
dp[n - 1] = arr[n - 1];
// Max-heap to store {dp[index], index}
let maxh = [];
maxh.push([dp[n - 1], n - 1]);
// Iterate from second last index to the first
for (let i = n - 2; i >= 0; i--) {
// Remove out-of-range elements from the heap
while (maxh.length && maxh[0][1] > i + k) {
maxh.shift();
}
// Set dp[i] as the max value from the valid
// range + current index value
dp[i] = maxh[0][0] + arr[i];
// Push current dp[i] and index into the heap
maxh.push([dp[i], i]);
// Sort heap to maintain max-heap property
maxh.sort((a, b) => b[0] - a[0]);
}
// Return the maximum score starting from the
// first index
return dp[0];
}
let arr = [100, -30, -50, -15, -20, -30];
let k = 3;
console.log(getScore(arr, k));
Using DP + Deque - O(n) Time and O(k) Space
The Deque (Double-Ended Queue) provides an efficient alternative to the max-heap by maintaining a sliding window of indices corresponding to maximum values in the dp array. Unlike a heap, which requires logarithmic operations, the deque allows for constant time operations to remove outdated indices and insert new ones, making it ideal for this problem.
Steps to implement the above idea:
- Initialize a dp array of size n with minimum integer value.
- Set dp[n-1] as arr[n-1] and push its index into a deque.
- Iterate backward from the second last index to the first.
- Remove out-of-range indices from the front of the deque.
- Set dp[i] as arr[i] + dp[dq.front()].
- Maintain deque order, remove smaller values from the back, and push I.
C++
// C++ implementation to find max score
// from at most k jumps using Tabulation + Deque
#include <bits/stdc++.h>
using namespace std;
// Function to calculate maximum score using
// Tabulation + Deque
int getScore(vector<int> &arr, int k) {
int n = arr.size();
// Tabulation array to store maximum scores
vector<int> dp(n, INT_MIN);
// Base case: score at the last index is the
// value at that index
dp[n - 1] = arr[n - 1];
// Deque to store indices of dp values in
// decreasing order
deque<int> dq;
dq.push_back(n - 1);
// Iterate from second last index to the first
for (int i = n - 2; i >= 0; i--) {
// Remove indices from the deque that are out of range
while (!dq.empty() && dq.front() > i + k) {
dq.pop_front();
}
// Set dp[i] as the maximum score from the deque
dp[i] = arr[i] + dp[dq.front()];
// Maintain the deque in decreasing order of dp values
while (!dq.empty() && dp[i] >= dp[dq.back()]) {
dq.pop_back();
}
// Add the current index to the deque
dq.push_back(i);
}
// Return the maximum score starting from the
// first index
return dp[0];
}
int main() {
vector<int> arr = {100, -30, -50, -15, -20, -30};
int k = 3;
cout << getScore(arr, k) << endl;
return 0;
}
// Java implementation to find max score
// from at most k jumps using Tabulation + Deque
import java.util.*;
class GfG {
static int getScore(int[] arr, int k) {
int n = arr.length;
// Tabulation array to store scores
int[] dp = new int[n];
// Base case: score at the last index
dp[n - 1] = arr[n - 1];
// Deque to store indices of dp values
// in decreasing order
Deque<Integer> dq = new LinkedList<>();
dq.offer(n - 1);
// Iterate from second last index to the first
for (int i = n - 2; i >= 0; i--) {
// Remove indices from the deque that are out of range
while (!dq.isEmpty() && dq.peekFirst() > i + k) {
dq.pollFirst();
}
// Set dp[i] as the maximum score from the deque
dp[i] = arr[i] + dp[dq.peekFirst()];
// Maintain the deque in decreasing order of dp values
while (!dq.isEmpty() && dp[i] >= dp[dq.peekLast()]) {
dq.pollLast();
}
// Add the current index to the deque
dq.offerLast(i);
}
// Return the maximum score starting
// from the first index
return dp[0];
}
public static void main(String[] args) {
int[] arr = {100, -30, -50, -15, -20, -30};
int k = 3;
System.out.println(getScore(arr, k));
}
}
# Python implementation to find max score
# from at most k jumps using Tabulation + Deque
# Main function to get maximum score
from collections import deque
def getScore(arr, k):
n = len(arr)
# Tabulation array to store scores
dp = [float('-inf')] * n
# Base case: score at the last index
dp[n - 1] = arr[n - 1]
# Deque to store indices of dp values
# in decreasing order
dq = deque([n - 1])
# Iterate from second last index to first
for i in range(n - 2, -1, -1):
# Remove indices from deque that are out of range
while dq and dq[0] > i + k:
dq.popleft()
# Set dp[i] as the maximum score from the deque
dp[i] = arr[i] + dp[dq[0]]
# Maintain deque in decreasing order of dp values
while dq and dp[i] >= dp[dq[-1]]:
dq.pop()
# Add current index to the deque
dq.append(i)
# Return score from the first index
return dp[0]
if __name__ == "__main__":
arr = [100, -30, -50, -15, -20, -30]
k = 3
print(getScore(arr, k))
// C# implementation to find max score
// from at most k jumps using Tabulation + Deque
using System;
using System.Collections.Generic;
class GfG {
static int GetScore(int[] arr, int k) {
int n = arr.Length;
// Tabulation array to store scores
int[] dp = new int[n];
// Base case: score at the last index
dp[n - 1] = arr[n - 1];
// Deque to store indices of dp values
// in decreasing order
LinkedList<int> dq = new LinkedList<int>();
dq.AddLast(n - 1);
// Iterate from second last index to the first
for (int i = n - 2; i >= 0; i--) {
// Remove indices from the deque that are out of range
while (dq.Count > 0 && dq.First.Value > i + k) {
dq.RemoveFirst();
}
// Set dp[i] as the maximum score from the deque
dp[i] = arr[i] + dp[dq.First.Value];
// Maintain the deque in decreasing order of dp values
while (dq.Count > 0 && dp[i] >= dp[dq.Last.Value]) {
dq.RemoveLast();
}
// Add the current index to the deque
dq.AddLast(i);
}
// Return the maximum score starting
// from the first index
return dp[0];
}
static void Main(string[] args) {
int[] arr = {100, -30, -50, -15, -20, -30};
int k = 3;
Console.WriteLine(GetScore(arr, k));
}
}
// JavaScript implementation to find max score
// from at most k jumps using Tabulation + Deque
function getScore(arr, k) {
let n = arr.length;
// Tabulation array to store scores
let dp = new Array(n).fill(-Infinity);
// Base case: score at the last index
dp[n - 1] = arr[n - 1];
// Deque to store indices of dp values
// in decreasing order
let dq = [];
// Add the last index to the deque
dq.push(n - 1);
// Iterate from second last index to first
for (let i = n - 2; i >= 0; i--) {
// Remove indices from deque that are out of range
while (dq.length > 0 && dq[0] > i + k) {
dq.shift();
}
// Set dp[i] as the maximum score from the deque
dp[i] = arr[i] + dp[dq[0]];
// Maintain deque in decreasing order of dp values
while (dq.length > 0 && dp[i] >= dp[dq[dq.length - 1]]) {
dq.pop();
}
// Add current index to the deque
dq.push(i);
}
// Return score from the first index
return dp[0];
}
//Driver code
let arr = [100, -30, -50, -15, -20, -30];
let k = 3;
console.log(getScore(arr, k));
Similar Reads
Maximize Sum possible from an Array by the given moves
Given three integers N, M and K and an array a[] consisting of N integers, where M and K denotes the total number of possible moves and the number of possible moves(shift by an index) on the left of the current element in an array respectively, the task is to maximize the sum possible by traversing
8 min read
Minimum cost to reach the end of the array with maximum jumps of length K
Given an array arr[] of size N and an integer K, one can move from an index i to any other index j such that j ? i+k. The cost of being at any index 'i', is arr[i]. The task is to find the minimum cost to reach the end of the array starting from index 0. Examples: Input : arr[] = {2, 4, 1, 6, 3}, K
10 min read
Maximize sum possible from an array by jumps of length i + K * arr[i] from any ith index
Given an array arr[] consisting of N positive integers and an integer K, the task is to find the maximum sum of array elements possible by jumps of (i + K*arr[i]) (< N) length from any ith index of the array. Examples: Input: arr[] = {1, 2, 1, 4}, K = 2Output: 4Explanation:Starting index i = 0, t
10 min read
Maximum frequency of any array element possible by at most K increments
Given an array arr[] of size N and an integer K, the task is to find the maximum possible frequency of any array element by at most K increments. Examples: Input: arr[] = {1, 4, 8, 13}, N = 4, K = 5 Output: 2 Explanation: Incrementing arr[0] twice modifies arr[] to {4, 4, 8, 13}. Maximum frequency =
15+ min read
Count maximum possible pairs from an array having sum K
Given an array arr[] consisting of N integers and an integer K, the task is to find the maximum number of pairs having a sum K possible from the given array. Note: Every array element can be part of a single pair. Examples: Input: arr[] = {1, 2, 3, 4}, K = 5Output: 2Explanation: Pairs with sum K fro
11 min read
Minimum length of X[] after removing at most K sub-arrays
Given an array X[] of length N and an integer K. Then your task is to output minimum length of X[] after removing at most K number of sub-arrays, which satisfies the following conditions: Size of the sub-array must be greater than 1.First and last element of sub-array must be same.Note: The sub-arra
10 min read
Maximum K-digit number possible from subsequences of two given arrays
Given two arrays arr1[] and arr2[] of length M and N consisting of digits [0, 9] representing two numbers and an integer K(K ? M + N), the task is to find the maximum K-digit number possible by selecting subsequences from the given arrays such that the relative order of the digits is the same as in
12 min read
Maximum possible number with concatenations of K numbers from given array
Given an array arr[] of N integers and a positive integer K, the task is to choose K integers from the array arr[] such that their concatenation of them forms the largest possible integer. All the array elements must be chosen at least once for creating the number. Note: It is always guaranteed that
6 min read
Minimize maximum array element possible by at most K splits on the given array
Given an array arr[] consisting of N positive integers and a positive integer K, the task is to minimize the maximum element present in the array by splitting at most K array elements into two numbers equal to their value. Examples: Input: arr[] = {2, 4, 8, 2}, K = 4Output: 2Explanation:Following se
9 min read
Collect maximum points in an array with k moves
Given an array of integer and two values k and i where k is the number of moves and i is the index in the array. The task is to collect maximum points in the array by moving either in single or both directions from given index i and making k moves. Note that every array element visited is considered
9 min read