Find minimum moves to reach target on an infinite line
Given a target position on infinite number line, i.e -infinity to +infinity. Starting form 0 you have to reach the target by moving as described : In ith move you can take i steps forward or backward. Find the minimum number of moves require to reach the target.
Input : target = 3 Output : 2 Explanation: On the first move we step from 0 to 1. On the second step we step from 1 to 3. Input: target = 2 Output: 3 Explanation: On the first move we step from 0 to 1. On the second move we step from 1 to -1. On the third move we step from -1 to 2.
We have discussed a naive recursive solution in below post.
Minimum steps to reach a destination
If target is negative, we can take it as positive because we start from 0 in symmetrical way.
Idea is to move in one direction as long as possible, this will give minimum moves. Starting at 0 first move takes us to 1, second move takes us to 3 (1+2) position, third move takes us to 6 (1+2+3) position, ans so on; So for finding target we keep on adding moves until we find the nth move such that 1+2+3+…+n>=target. Now if sum (1+2+3+…+n) is equal to target the our job is done, i.e we’ll need n moves to reach target. Now next case where sum is greater than target. Find the difference by how much we are ahead, i.e sum – target. Let the difference be d = sum – target.
Example: if the target is 2 then, the number of steps should be 3. Lets see how we get that,
1-2+3=2 ( if we start from 0 and move 1 step forward and then 2 steps backward (we land on -1) and further if take 3 steps forward we land on 2.)
1-2+3 can also be written as (1+2+3) -2*(2).
that means if we take the ith move backward then the new sum will become (sum – 2i).
Now if sum-2i = target then our job is done. Since, sum – target = 2i, i.e difference should be even as we will get an integer i flipping which will give the answer. So following cases arise.
Case 1 : Difference is even then answer is n, (because we will always get a move flipping which will lead to target).
Case 2 : Difference is odd, then we take one more step, i.e add n+1 to sum and now again take the difference. If difference is even the n+1 is the answer else we would have to take one more move and this will certainly make the difference even then answer will be n+2.
Explanation : Since difference is odd. Target is either odd or even.
case 1: n is even (1+2+3+…+n) then adding n+1 makes the difference even.
case 2: n is odd then adding n+1 doesn’t makes difference even so we would have to take one more move, so n+2.
target = 5.
we keep on taking moves until we reach target or we just cross it.
sum = 1 + 2 + 3 = 6 > 5, step = 3.
Difference = 6 – 5 = 1. Since the difference is an odd value, we will not reach the target by flipping any move from +i to -i. So we increase our step. We need to increase step by 2 to get an even difference (since n is odd and target is also odd). Now that we have an even difference, we can simply switch any move to the left (i.e. change + to -) as long as the summation of the changed value equals to half of the difference. We can switch 1 and 4 or 2 and 3 or 5.
Time Complexity: O(n), where n is the number of steps
Auxiliary Space: O(1), since no extra space has been taken.
Please suggest if someone has a better solution which is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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