# Solution

• I have created two solutions for this problem, the first one was finding the fewest steps that's required to reach the destination (end of the array) for each nodes, which would solve this problem. However, I wasn't able to submit that one because it would fail at one test case which has many many nodes with execution time exceeded.

This is the second solution I created, it is pretty simple, which simply finds the quickest path to the destination from any given node (in this case node 0). And here's the solution:

``````class Solution {
public:
int jump(vector<int>& nums) {
int currentIndex = 0;
int currentStepCount = 0;

if(nums.size() == 1)
{
// A shortcut for optimization purpose
return 0;
}

while(true)
{
if(currentIndex >= nums.size())
{
// We have already reached or passed our destination
break;
}
else if(currentIndex + nums[currentIndex] >= nums.size()-1)
{
// We know we will reach our destination in the next move
currentStepCount++;
break;
}

currentIndex = currentMaximumReach(nums, currentIndex);
currentStepCount++;
}

return currentStepCount;
}

int currentMaximumReach(vector<int> nums, int index)
{
//
// This function returns the index of the next node that will give the maximum reach from the current node
//

int MaxReach = index+nums[index];
int MaxReachNode = index;

for(int i = 1; (i+index < nums.size()) && (i <= nums[index]); i++)
{
if(nums[index+i]+index+i > MaxReach)
{
MaxReach = index+i+nums[index+i];
MaxReachNode = index+i;
}
}

// Return the node that would give the max reach from the current node
return MaxReachNode;
}
};
``````

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