# 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.**Examples :**

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.

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the **Essential Maths for CP Course** at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

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.

If we take the i-th move backward then the new sum will become (sum – 2i), i.e 1+2+3+…-x+x+1…+n. 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.**Example: **

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.

## C++

`// CPP program to find minimum moves to` `// reach target if we can move i steps in` `// i-th move.` `#include <iostream>` `using` `namespace` `std;` `int` `reachTarget(` `int` `target)` `{` ` ` `// Handling negatives by symmetry` ` ` `target = ` `abs` `(target);` ` ` ` ` `// Keep moving while sum is smaller or difference` ` ` `// is odd.` ` ` `int` `sum = 0, step = 0;` ` ` `while` `(sum < target || (sum - target) % 2 != 0) {` ` ` `step++;` ` ` `sum += step;` ` ` `}` ` ` `return` `step;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `target = 5;` ` ` `cout << reachTarget(target);` ` ` `return` `0;` `}` |

## Java

`// Java program to find minimum ` `//moves to reach target if we can` `// move i steps in i-th move.` `import` `java.io.*;` `import` `java.math.*;` `class` `GFG {` ` ` ` ` `static` `int` `reachTarget(` `int` `target)` ` ` `{` ` ` `// Handling negatives by symmetry` ` ` `target = Math.abs(target);` ` ` ` ` `// Keep moving while sum is smaller` ` ` `// or difference is odd.` ` ` `int` `sum = ` `0` `, step = ` `0` `;` ` ` ` ` `while` `(sum < target || (sum - target) % ` `2` ` ` `!= ` `0` `) {` ` ` `step++;` ` ` `sum += step;` ` ` `}` ` ` `return` `step;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `int` `target = ` `5` `;` ` ` `System.out.println(reachTarget(target));` ` ` `}` `}` `// This code is contributed by Nikita tiwari.` |

## Python 3

`# Python 3 program to find minimum ` `# moves to reach target if we can` `# move i steps in i-th move.` `def` `reachTarget(target) :` ` ` `# Handling negatives by symmetry` ` ` `target ` `=` `abs` `(target)` ` ` ` ` `# Keep moving while sum is` ` ` `# smaller or difference is odd.` ` ` `sum` `=` `0` ` ` `step ` `=` `0` ` ` `while` `(` `sum` `< target ` `or` `(` `sum` `-` `target) ` `%` ` ` `2` `!` `=` `0` `) :` ` ` `step ` `=` `step ` `+` `1` ` ` `sum` `=` `sum` `+` `step` ` ` ` ` `return` `step` ` ` `# Driver code` `target ` `=` `5` `print` `(reachTarget(target))` `# This code is contributed by Nikita Tiwari` |

## C#

`// C# program to find minimum` `//moves to reach target if we can` `// move i steps in i-th move.` `using` `System;` `class` `GFG {` ` ` ` ` `static` `int` `reachTarget(` `int` `target)` ` ` `{` ` ` `// Handling negatives by symmetry` ` ` `target = Math.Abs(target);` ` ` ` ` `// Keep moving while sum is smaller` ` ` `// or difference is odd.` ` ` `int` `sum = 0, step = 0;` ` ` ` ` `while` `(sum < target ||` ` ` `(sum - target) % 2!= 0)` ` ` `{` ` ` `step++;` ` ` `sum += step;` ` ` `}` ` ` `return` `step;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `target = 5;` ` ` `Console.WriteLine(reachTarget(target));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## PHP

`<?php` `// PHP program to find ` `// minimum moves to reach` `// target if we can move i` `// steps in i-th move.` `function` `reachTarget(` `$target` `)` `{` ` ` `// Handling negatives` ` ` `// by symmetry` ` ` `$target` `= ` `abs` `(` `$target` `);` ` ` ` ` `// Keep moving while sum is` ` ` `// smaller or difference is odd.` ` ` `$sum` `= 0; ` `$step` `= 0;` ` ` `while` `(` `$sum` `< ` `$target` `or` ` ` `(` `$sum` `- ` `$target` `) % 2 != 0)` ` ` `{` ` ` `$step` `++;` ` ` `$sum` `+= ` `$step` `;` ` ` `}` ` ` `return` `$step` `;` `}` `// Driver code` `$target` `= 5;` `echo` `reachTarget(` `$target` `);` `// This code is contributed by anuj_67.` `?>` |

## Javascript

`<script>` `// JavaScript program to find minimum ` `//moves to reach target if we can` `// move i steps in i-th move.` `function` `reachTarget(target)` ` ` `{` ` ` `// Handling negatives by symmetry` ` ` `target = Math.abs(target);` ` ` ` ` `// Keep moving while sum is smaller` ` ` `// or difference is odd.` ` ` `let sum = 0, step = 0;` ` ` ` ` `while` `(sum < target || (sum - target) % 2` ` ` `!= 0) {` ` ` `step++;` ` ` `sum += step;` ` ` `}` ` ` `return` `step;` ` ` `}` `// Driver code` ` ` `let target = 5;` ` ` `document.write(reachTarget(target));` `</script>` |

**Output :**

5